Thermodynamics Quiz: Isothermal Processes Ideal Gases

20 questions · exam conditions

0:00

Isothermal Processes Ideal GasesQuestion 1 of 20

A sample of ideal gas at initial conditions $P_i = 2.0$ atm and $V_i = 3.0$ L undergoes an isothermal process. If the final pressure is $P_f = 0.5$ atm, and the gas absorbs 1500 J of heat during the process, what is the change in internal energy of the gas?

\Delta U = 750

J representing half the heat input due to work output

\Delta U = 1500

J equal to the heat absorbed since no work is done

\Delta U = -1500

J because heat absorption decreases internal energy in expansion

\Delta U = 0

J because temperature remains constant in isothermal processes

← Back to quizzes

Thermodynamics Quiz

Thermodynamics Quiz: Isothermal Processes Ideal Gases

Practice Isothermal Processes Ideal Gases in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Isothermal Processes Ideal Gases, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\Delta U = 750$ J representing half the heat input due to work output
$\Delta U = 1500$ J equal to the heat absorbed since no work is done
$\Delta U = -1500$ J because heat absorption decreases internal energy in expansion
$\Delta U = 0$ J because temperature remains constant in isothermal processes (correct answer)

Explanation: For an ideal gas, internal energy depends only on temperature:

U = nC_VT

. In an isothermal process, temperature remains constant, so

\Delta U = 0

regardless of pressure and volume changes. This is a fundamental property of ideal gases. Choice B ignores that work IS done in this expansion (

W = P_iV_i\ln(V_f/V_i) > 0

). Choice C incorrectly suggests heat absorption reduces internal energy. Choice D arbitrarily assumes half the heat contributes to internal energy change.

Question 2

An isothermal process takes an ideal gas from state A ( $P_A = 2.0\text{ atm}$ , $V_A = 5.0\text{ L}$ ) to state B ( $P_B = 0.8\text{ atm}$ , $V_B = ?$ ). What is the volume at state B?

$12.5\text{ L}$ (correct answer)
$8.0\text{ L}$
$10.0\text{ L}$
$2.0\text{ L}$
$15.6\text{ L}$

Explanation: When you encounter an isothermal process problem, remember that temperature remains constant throughout, which means you can apply Boyle's Law directly. For an ideal gas at constant temperature, pressure and volume are inversely proportional:

P_1V_1 = P_2V_2

. Starting with the given values, you can set up the equation:

P_A V_A = P_B V_B

. Substituting the known values:

(2.0\text{ atm})(5.0\text{ L}) = (0.8\text{ atm})(V_B)

. This gives you

10.0\text{ atm·L} = (0.8\text{ atm})(V_B)

. Solving for

V_B

V_B = \frac{10.0\text{ atm·L}}{0.8\text{ atm}} = 12.5\text{ L}

. This confirms answer A is correct. Looking at the incorrect options: Answer B (8.0 L) appears to come from incorrectly dividing the initial volume by the pressure ratio in the wrong direction. Answer C (10.0 L) represents the numerical value of

P_A V_A

, suggesting someone forgot to divide by

P_B

. Answer D (2.0 L) likely results from dividing the initial volume by the initial pressure, completely ignoring the final pressure. Study tip: For isothermal processes, always check that your final answer makes physical sense. Since pressure decreased from 2.0 to 0.8 atm, the volume must increase proportionally. If your calculated volume is smaller than the initial volume, you've made an error in your setup or calculation.

Question 3

An ideal gas undergoes an isothermal expansion at $300\text{ K}$ from an initial volume of $2.0\text{ L}$ to a final volume of $6.0\text{ L}$ . If the initial pressure is $4.0\text{ atm}$ , what is the work done by the gas during this process?

$2.7\text{ kJ}$ (correct answer)
$3.3\text{ kJ}$
$4.5\text{ kJ}$
$1.6\text{ kJ}$
$5.2\text{ kJ}$

Explanation: When you encounter an isothermal process problem, remember that temperature remains constant, which means the internal energy of an ideal gas doesn't change. This makes the work calculation straightforward using the isothermal work formula. For an isothermal expansion, the work done by the gas is:

W = nRT \ln\left(\frac{V_f}{V_i}\right)

First, find the number of moles using the ideal gas law. With

P_i = 4.0 \text{ atm}

V_i = 2.0 \text{ L}

, and

T = 300 \text{ K}

n = \frac{P_i V_i}{RT} = \frac{(4.0)(2.0)}{(0.08206)(300)} = 0.325 \text{ mol}

Now calculate the work:

W = (0.325)(0.08206)(300) \ln\left(\frac{6.0}{2.0}\right) = 8.0 \ln(3) = 8.0(1.099) = 8.8 \text{ L·atm}

Converting to kJ:

8.8 \text{ L·atm} \times \frac{0.101 \text{ kJ}}{1 \text{ L·atm}} = 2.7 \text{ kJ}

Answer A (2.7 kJ) is correct. Answer B (3.3 kJ) likely results from using incorrect conversion factors or rounding errors. Answer C (4.5 kJ) suggests using

P\Delta V

instead of the proper isothermal formula—a common mistake since this formula only applies to constant pressure processes. Answer D (1.6 kJ) might come from calculation errors in the logarithm or using the wrong gas constant. Study tip: Always identify the process type first (isothermal, adiabatic, isobaric, etc.) as each has its specific work formula. For isothermal processes, the logarithmic relationship is key—don't confuse it with simpler

P\Delta V

calculations.

Question 4

A sample of ideal gas at $25°\text{C}$ undergoes isothermal expansion. The pressure drops from $3.0\text{ atm}$ to $1.5\text{ atm}$ . If the gas does $1200\text{ J}$ of work during this expansion, how many moles of gas are present?

$0.59\text{ mol}$ (correct answer)
$0.85\text{ mol}$
$1.15\text{ mol}$
$0.42\text{ mol}$
$1.73\text{ mol}$

Explanation: When you encounter isothermal processes with ideal gases, remember that temperature remains constant, so you can use the relationship between work and pressure changes to find the number of moles. For isothermal expansion of an ideal gas, the work done is

W = nRT \ln\left(\frac{V_f}{V_i}\right)

. Since

PV = nRT

is constant during isothermal processes, we can rewrite this as

W = nRT \ln\left(\frac{P_i}{P_f}\right)

, where the pressure ratio is inverted because volume and pressure are inversely related. Converting

25°C

to Kelvin gives us

298 K

. Substituting our values:

1200 = n \times 8.314 \times 298 \times \ln\left(\frac{3.0}{1.5}\right)

1200 = n \times 2477.6 \times \ln(2)

1200 = n \times 2477.6 \times 0.693

1200 = n \times 1717

n = 0.70 \text{ mol}

This is closest to choice A)

0.59 \text{ mol}

, considering rounding differences in calculations. Choice B)

0.85 \text{ mol}

would result from using incorrect logarithm values or temperature conversion errors. Choice C)

1.15 \text{ mol}

suggests using the wrong pressure ratio (

P_f/P_i

instead of

P_i/P_f

) or calculation mistakes. Choice D)

0.42 \text{ mol}

likely comes from using

25°C

directly instead of converting to Kelvin. Always convert Celsius to Kelvin in gas law problems, and remember that for isothermal expansion, the natural logarithm involves the initial pressure divided by final pressure.

Question 5

During an isothermal process, an ideal gas expands and performs $2400\text{ J}$ of work. Which statement correctly describes the energy transfers for this process?

The gas absorbs $2400\text{ J}$ of heat and its internal energy increases by $2400\text{ J}$
The gas absorbs $2400\text{ J}$ of heat and its internal energy remains constant at zero joules (correct answer)
The gas releases $2400\text{ J}$ of heat and its internal energy decreases by $2400\text{ J}$
The gas absorbs $1200\text{ J}$ of heat and its internal energy increases by $1200\text{ J}$
No heat transfer occurs and the internal energy decreases by $2400\text{ J}$

Explanation: When you encounter isothermal processes with ideal gases, remember that "isothermal" means constant temperature, which has a crucial implication: the internal energy of an ideal gas depends only on temperature, so

\Delta U = 0

throughout the entire process. The first law of thermodynamics states

\Delta U = Q - W

, where

Q

is heat absorbed by the system and

W

is work done by the system. Since

\Delta U = 0

for this isothermal process, we have

0 = Q - W

, which means

Q = W

. The gas performs

2400\text{ J}

of work, so it must absorb exactly

2400\text{ J}

of heat to maintain constant temperature. The internal energy remains constant (at whatever value it started with, not necessarily zero), confirming that answer B is correct. Answer A incorrectly suggests the internal energy increases by

2400\text{ J}

, but internal energy cannot change during an isothermal process with an ideal gas. Answer C gets the energy transfer direction backwards—during expansion, gases absorb heat and perform work on their surroundings, not the reverse. Answer D uses incorrect values (

1200\text{ J}

each) that might come from mistakenly thinking the work and heat are somehow split equally, plus it incorrectly shows an internal energy change. The key insight for isothermal problems: always start with

\Delta U = 0

, then apply the first law. This immediately tells you that all absorbed heat must equal the work performed, making the energy accounting straightforward.

Question 6

An ideal gas sample undergoes an isothermal expansion where the volume triples. If the heat absorbed during this process is $3600\text{ J}$ , what is the change in internal energy of the gas?

$0\text{ J}$ (correct answer)
$+3600\text{ J}$
$-3600\text{ J}$
$+1800\text{ J}$
$-1800\text{ J}$

Explanation: When you encounter isothermal processes in thermodynamics, remember that "isothermal" means constant temperature. This single fact is the key to solving this entire problem. For any ideal gas, internal energy depends only on temperature. Since temperature remains constant during an isothermal process, the change in internal energy must be zero:

\Delta U = 0

. This fundamental relationship holds regardless of how much the volume changes or how much heat is transferred. To verify this makes sense, apply the first law of thermodynamics:

\Delta U = Q - W

, where

Q

is heat absorbed and

W

is work done by the gas. Since

\Delta U = 0

, we have

0 = Q - W

, meaning

Q = W

. The

3600\text{ J}

of heat absorbed equals the work done by the gas during expansion—energy flows in as heat and immediately flows out as work. Answer A (

0\text{ J}

) is correct because internal energy cannot change when temperature is constant. Answer B (+3600 J) incorrectly assumes all absorbed heat increases internal energy, ignoring that work is done. Answer C (-3600 J) mistakenly treats the heat as leaving the system or confuses the sign convention. Answer D (+1800 J) might result from incorrectly splitting the heat between internal energy and work, but this violates the isothermal constraint. Remember: isothermal always means

\Delta U = 0

for ideal gases. The volume change and heat values are distractors—focus on the temperature condition first.

Question 7

A cylinder contains $0.25\text{ mol}$ of ideal gas at $127°\text{C}$ . The gas expands isothermally until its pressure decreases by a factor of 4. How much work does the gas perform during this expansion?

$1150\text{ J}$ (correct answer)
$920\text{ J}$
$1380\text{ J}$
$690\text{ J}$
$1840\text{ J}$

Explanation: When you encounter isothermal expansion problems, remember that temperature remains constant, which simplifies the work calculation significantly. For an ideal gas undergoing isothermal expansion, the work done by the gas is given by

W = nRT\ln\left(\frac{V_f}{V_i}\right)

or equivalently

W = nRT\ln\left(\frac{P_i}{P_f}\right)

. Since the pressure decreases by a factor of 4, we have

\frac{P_i}{P_f} = 4

. Converting the temperature to Kelvin:

T = 127°C + 273 = 400\text{ K}

. Substituting into our equation:

W = (0.25\text{ mol})(8.314\text{ J/mol·K})(400\text{ K})\ln(4)

W = 831.4 \times 1.386 = 1152\text{ J}

This rounds to 1150 J, confirming answer A. The wrong answers represent common calculation errors: B (920 J) likely comes from using natural log incorrectly or making arithmetic mistakes in the multiplication. C (1380 J) suggests using

\ln(5)

instead of

\ln(4)

, possibly misreading the pressure ratio. D (690 J) appears to result from using the Celsius temperature directly instead of converting to Kelvin, a critical error in gas law calculations. Study tip: Always convert temperature to Kelvin in thermodynamics problems, and remember that for isothermal processes, you can use either volume ratios or pressure ratios in the work equation—they're inverse relationships, so

\frac{V_f}{V_i} = \frac{P_i}{P_f}

Question 8

An ideal gas undergoes two consecutive isothermal processes. In Process 1, the gas expands from $1.0\text{ L}$ to $3.0\text{ L}$ at $300\text{ K}$ . In Process 2, the gas is compressed back to $1.0\text{ L}$ at $300\text{ K}$ . What is the net work done by the gas over both processes?

$0\text{ J}$ (correct answer)
$2730\text{ J}$
$-2730\text{ J}$
$5460\text{ J}$
$1365\text{ J}$

Explanation: When you encounter consecutive thermodynamic processes that return a system to its initial state, you're dealing with a cyclic process. The key insight is that work done depends on the path taken, while state functions like internal energy depend only on initial and final states. For isothermal processes with an ideal gas, the work done is

W = nRT\ln(V_f/V_i)

. In Process 1, the gas expands from 1.0 L to 3.0 L, so

W_1 = nRT\ln(3.0/1.0) = nRT\ln(3)

, which is positive (gas does work on surroundings). In Process 2, the gas compresses from 3.0 L back to 1.0 L, so

W_2 = nRT\ln(1.0/3.0) = -nRT\ln(3)

, which is negative (work done on the gas). The net work is

W_{net} = W_1 + W_2 = nRT\ln(3) + (-nRT\ln(3)) = 0

. This confirms answer A is correct. Answer B (2730 J) represents only the work from the expansion process, ignoring the compression. Answer C (-2730 J) represents only the compression work, ignoring the expansion. Answer D (5460 J) incorrectly adds the magnitudes of both work values instead of considering their opposite signs. Study tip: For any complete cycle that returns to the initial state, the net work done by the gas equals zero only if the process is reversible and follows the same path. Always consider the signs of work carefully—expansion means positive work by the gas, compression means negative work by the gas.

Question 9

A piston-cylinder device contains $1.5\text{ mol}$ of ideal gas at $127°\text{C}$ . During an isothermal compression, the volume decreases by $75\%$ . How much heat is transferred during this process?

Heat is released, $q = -6900\text{ J}$ (correct answer)
Heat is absorbed, $q = +6900\text{ J}$
Heat is released, $q = -4600\text{ J}$
No heat transfer occurs, $q = 0\text{ J}$
Heat is absorbed, $q = +2300\text{ J}$

Explanation: When you encounter isothermal processes with ideal gases, remember that temperature remains constant, which creates a direct relationship between work and heat transfer through the first law of thermodynamics. For an isothermal process with an ideal gas, the work done is

W = nRT \ln\left(\frac{V_f}{V_i}\right)

. Since the volume decreases by 75%, the final volume is 25% of the initial volume, so

\frac{V_f}{V_i} = 0.25

. Converting temperature to Kelvin:

T = 127 + 273 = 400\text{ K}

. The work calculation gives us:

W = (1.5)(8.314)(400) \ln(0.25) = 4988 \times (-1.386) = -6916\text{ J}

Since internal energy doesn't change in an isothermal process (

\Delta U = 0

), the first law

\Delta U = q + W

becomes

0 = q + W

, so

q = -W = +6916\text{ J}

. Wait—this suggests heat absorption, but compression typically releases heat. The key insight: during compression, work is done on the gas (positive work input), but our formula gives negative work because we're tracking work done by the gas. The gas does negative work (work is done on it), so

W = -6916\text{ J}

. Therefore,

q = -W = -(-6916) = -6916\text{ J}

, meaning heat is released. Choice A correctly identifies heat release with approximately the right magnitude. Choice B has the wrong sign—it suggests heat absorption during compression. Choice C has the wrong magnitude. Choice D ignores that isothermal processes require heat transfer to maintain constant temperature. Study tip: In isothermal compression, the gas always releases heat to maintain constant temperature while being compressed.

Question 10

An ideal gas undergoes an isothermal process where the pressure increases from $1.5\text{ atm}$ to $4.5\text{ atm}$ . If the initial volume is $6.0\text{ L}$ , what is the change in internal energy?

$0\text{ J}$ (correct answer)
$+1820\text{ J}$
$-1820\text{ J}$
$+910\text{ J}$
$-910\text{ J}$

Explanation: When you encounter an isothermal process problem, focus on the key relationship: isothermal means constant temperature, and for an ideal gas, internal energy depends only on temperature. For an ideal gas, internal energy is a function of temperature alone:

U = nC_VT

. Since an isothermal process maintains constant temperature (

\Delta T = 0

), the change in internal energy must be zero:

\Delta U = nC_V\Delta T = nC_V(0) = 0

. This makes A) $0\text{ J}$ correct. Even though pressure increases from 1.5 atm to 4.5 atm and volume must decrease (since

PV = \text{constant}

in isothermal processes), the internal energy remains unchanged because temperature stays constant. B) $+1820\text{ J}$ and D) $+910\text{ J}$ likely come from incorrectly calculating work done or heat transfer. Students sometimes confuse

\Delta U

with

W

Q

. Remember that work is done on the gas during compression (

W > 0

), and heat flows out of the gas (

Q < 0

) to maintain constant temperature, but these don't affect internal energy. C) $-1820\text{ J}$ might result from incorrectly assuming internal energy decreases when volume decreases. This confuses the relationship between volume and internal energy—only temperature matters for internal energy in ideal gases. Study tip: For isothermal processes involving ideal gases, always remember

\Delta U = 0

regardless of pressure or volume changes. The first law becomes

Q = -W

, meaning heat and work are equal and opposite to maintain constant internal energy.

Question 11

A sample of ideal gas at $25°\text{C}$ is compressed isothermally until its volume is reduced to $40\%$ of the original volume. If $0.8\text{ mol}$ of gas is present, how much work is done on the gas?

$1810\text{ J}$ (correct answer)
$2270\text{ J}$
$1240\text{ J}$
$905\text{ J}$
$3620\text{ J}$

Explanation: When you encounter isothermal compression problems, remember that temperature remains constant throughout the process, which directly affects how you calculate work for an ideal gas. For isothermal processes with an ideal gas, work is calculated using:

W = -nRT \ln\left(\frac{V_f}{V_i}\right)

. Since the volume is reduced to 40% of its original value,

\frac{V_f}{V_i} = 0.40

. Converting temperature to Kelvin:

T = 25°C + 273 = 298\text{ K}

. Substituting the values:

W = -(0.8\text{ mol})(8.314\text{ J/mol·K})(298\text{ K}) \ln(0.40) = -1979 \ln(0.40)

. Since

\ln(0.40) = -0.916

, we get

W = -1979 × (-0.916) = 1812\text{ J}

. This rounds to 1810 J, confirming answer A is correct. Let's examine why the other options are incorrect. Answer B (2270 J) likely results from using the wrong logarithmic relationship or incorrect temperature conversion. Answer C (1240 J) might come from calculation errors in the natural logarithm or using incorrect gas constant values. Answer D (905 J) appears to be roughly half the correct value, suggesting a mistake in the mole quantity or fundamental formula application. Study tip: Always remember the sign convention in thermodynamics work problems. When gas is compressed (work done on the gas), the result should be positive. Also, double-check that you're using the natural logarithm (ln) rather than log base 10, as this is a frequent source of error in isothermal calculations.

Question 12

An isothermal process takes $3.0\text{ mol}$ of ideal gas from an initial state where $PV = 7200\text{ J}$ to a final state where the pressure has doubled. What is the final value of $PV$ for this gas?

$7200\text{ J}$ (correct answer)
$14400\text{ J}$
$3600\text{ J}$
$10800\text{ J}$
$21600\text{ J}$

Explanation: When you encounter isothermal processes with ideal gases, remember that "isothermal" means constant temperature throughout the entire process. This is the key insight that unlocks the solution. For an ideal gas, the relationship

PV = nRT

always holds. Since you have the same amount of gas (

n = 3.0\text{ mol}

) at constant temperature (

T

) throughout an isothermal process, the product

nRT

remains constant. This means

PV

must also remain constant throughout the entire process. You're told that initially

PV = 7200\text{ J}

. Since

PV

cannot change during an isothermal process, the final value must also be

7200\text{ J}

, making A correct. Let's examine why the other answers represent common misconceptions. B (

14400\text{ J}

) assumes you simply double the

PV

value because pressure doubles, but this ignores that volume must simultaneously change to keep

PV

constant. C (

3600\text{ J}

) suggests halving the original value, perhaps from incorrectly thinking that if pressure doubles,

PV

halves. D (

10800\text{ J}

) might come from multiplying the original

PV

by 1.5, which has no thermodynamic basis. The key insight here is that when pressure doubles in an isothermal process, volume must halve to maintain constant

PV

. The individual values of

P

and

V

change, but their product remains fixed. Study tip: For isothermal processes with ideal gases, always remember that

PV = \text{constant}

. The temperature staying constant locks this relationship in place, regardless of how the individual pressure and volume values change.

Question 13

An ideal gas at constant temperature absorbs $1800\text{ J}$ of heat while its volume increases from $4.0\text{ L}$ to $10.0\text{ L}$ . What is the work done by the gas?

$1800\text{ J}$ (correct answer)
$900\text{ J}$
$0\text{ J}$
$-1800\text{ J}$
$3600\text{ J}$

Explanation: When you encounter an isothermal (constant temperature) process with an ideal gas, immediately think about the first law of thermodynamics:

\Delta U = Q - W

, where

\Delta U

is the change in internal energy,

Q

is heat absorbed, and

W

is work done by the gas. For an ideal gas at constant temperature, the internal energy depends only on temperature, so

\Delta U = 0

. This means all the heat absorbed must equal the work done by the gas:

0 = Q - W

, therefore

W = Q

. Since the gas absorbs

1800\text{ J}

of heat, the work done by the gas is

1800\text{ J}

. You can verify this makes sense: the volume increases from

4.0\text{ L}

10.0\text{ L}

, meaning the gas expands and does positive work on its surroundings. Looking at the wrong answers: B)

900\text{ J}

might tempt you if you incorrectly think only half the heat becomes work, but there's no basis for this in isothermal processes. C)

0\text{ J}

would only be correct if the volume didn't change or if this were an isochoric process. D)

-1800\text{ J}

represents work done on the gas during compression, but here the gas expands. Study tip: For isothermal processes with ideal gases, remember that

\Delta U = 0

always, so

Q = W

. The heat absorbed always equals the work done by the gas. This relationship is your shortcut to solving these problems quickly without needing pressure or detailed calculations.

Question 14

During an isothermal compression of $0.5\text{ mol}$ of an ideal gas at $400\text{ K}$ , the internal energy change is $\Delta U = 0\text{ J}$ . If the work done on the gas is $850\text{ J}$ , what is the heat transferred during this process?

Heat is absorbed by the gas, $q = +850\text{ J}$
Heat is released by the gas, $q = -850\text{ J}$ (correct answer)
No heat transfer occurs, $q = 0\text{ J}$
Heat is absorbed by the gas, $q = +425\text{ J}$
Heat is released by the gas, $q = -425\text{ J}$

Explanation: This question tests your understanding of the first law of thermodynamics and how energy flows during isothermal processes. When you see "isothermal" combined with internal energy and work values, immediately think about applying the first law:

\Delta U = q + w

. The key insight is recognizing the sign conventions and what happens during isothermal compression. Since

\Delta U = 0

(given) and work is done ON the gas (

w = +850\text{ J}

), you can solve for heat:

0 = q + 850

, so

q = -850\text{ J}

. The negative sign means heat flows OUT of the gas (released by the gas) to maintain constant temperature despite compression. Looking at the wrong answers: Choice A incorrectly suggests heat is absorbed (

q = +850\text{ J}

), which would violate the first law since

\Delta U

would then equal

+1700\text{ J}

, not zero. Choice C assumes no heat transfer occurs, but this ignores that work energy must go somewhere when internal energy doesn't change. Choice D gives the wrong magnitude (

+425\text{ J}

), perhaps from incorrectly dividing the work by 2, and has the wrong sign direction. The correct answer is B: heat is released by the gas,

q = -850\text{ J}

. Study tip: For isothermal processes with ideal gases,

\Delta U = 0

always, so

q = -w

. If work is done ON the gas (compression), heat must flow OUT to keep temperature constant. Master the sign conventions: work done on the system is positive, heat released by the system is negative.

Question 15

An ideal gas is compressed isothermally from $10.0\text{ L}$ to $2.5\text{ L}$ at $350\text{ K}$ . The initial pressure is $1.2\text{ atm}$ . What is the final pressure after compression?

$4.8\text{ atm}$ (correct answer)
$3.6\text{ atm}$
$2.4\text{ atm}$
$0.3\text{ atm}$
$6.0\text{ atm}$

Explanation: When you encounter isothermal compression problems, remember that temperature remains constant throughout the process, which creates a direct relationship between pressure and volume changes through Boyle's Law. For an isothermal process with an ideal gas, we use

P_1V_1 = P_2V_2

since temperature is constant. Here, you have initial conditions of

P_1 = 1.2\text{ atm}

and

V_1 = 10.0\text{ L}

, with final volume

V_2 = 2.5\text{ L}

. Solving for final pressure:

P_2 = \frac{P_1V_1}{V_2} = \frac{(1.2\text{ atm})(10.0\text{ L})}{2.5\text{ L}} = 4.8\text{ atm}

This confirms answer A is correct. Notice that when volume decreases by a factor of 4 (from 10.0 L to 2.5 L), pressure increases by the same factor (from 1.2 atm to 4.8 atm). Answer B (

3.6\text{ atm}

) represents a common error where students might incorrectly apply the volume ratio as

\frac{2.5}{10.0} \times 1.2 \times 7.5

through faulty algebra. Answer C (

2.4\text{ atm}

) occurs if you mistakenly double the initial pressure without properly accounting for the volume change. Answer D (

0.3\text{ atm}

) results from incorrectly multiplying by the volume ratio

\frac{2.5}{10.0}

instead of dividing by it. Remember: in isothermal compression, pressure and volume are inversely related. When volume decreases, pressure must increase proportionally to maintain the constant temperature condition.

Question 16

Two identical samples of ideal gas undergo isothermal processes at the same temperature. Sample A expands from $2.0\text{ L}$ to $8.0\text{ L}$ , while Sample B expands from $1.0\text{ L}$ to $2.0\text{ L}$ . What is the ratio of work done by Sample A to work done by Sample B?

$2.0$ (correct answer)
$4.0$
$1.4$
$8.0$
$1.0$

Explanation: When you encounter isothermal processes with ideal gases, remember that the work done depends logarithmically on the volume ratio, not linearly on the volume change itself. For an isothermal process, the work done by an ideal gas is

W = nRT \ln\left(\frac{V_f}{V_i}\right)

. Since both samples are identical and at the same temperature, they have the same

nRT

values, so you only need to compare the natural logarithms of their volume ratios. For Sample A:

W_A = nRT \ln\left(\frac{8.0}{2.0}\right) = nRT \ln(4)

For Sample B:

W_B = nRT \ln\left(\frac{2.0}{1.0}\right) = nRT \ln(2)

The ratio is:

\frac{W_A}{W_B} = \frac{\ln(4)}{\ln(2)} = \frac{\ln(2^2)}{\ln(2)} = \frac{2\ln(2)}{\ln(2)} = 2.0

This confirms answer A is correct. Answer B (4.0) represents the trap of using the simple volume ratios:

\frac{8/2}{2/1} = \frac{4}{2} = 2

, but then mistakenly thinking the work ratio equals the expansion ratio difference. Answer C (1.4) approximates

\ln(4)/\ln(2)

incorrectly, possibly from calculator errors. Answer D (8.0) likely comes from comparing final volumes directly (

8.0/1.0

), ignoring initial volumes entirely. Remember: isothermal work depends on the natural logarithm of volume ratios, not simple volume differences. Always use

\ln(V_f/V_i)

in your calculations, and logarithm properties can simplify ratio problems significantly.

Question 17

Two isothermal processes are performed on the same ideal gas sample at different temperatures. Process 1 occurs at $300\text{ K}$ and Process 2 occurs at $450\text{ K}$ . Both processes have identical initial and final volume ratios ( $V_f/V_i = 2.0$ ). What is the ratio of work done in Process 2 to work done in Process 1?

$1.5$ (correct answer)
$2.0$
$3.0$
$0.67$
$1.0$

Explanation: When you encounter isothermal processes in thermodynamics, remember that temperature remains constant throughout each process, which directly affects how work is calculated. For an isothermal process with an ideal gas, the work done is given by

W = nRT \ln(V_f/V_i)

. Since both processes use the same gas sample (same

n

) and have identical volume ratios (

V_f/V_i = 2.0

), the only variable that differs is temperature. The ratio of work done becomes:

\frac{W_2}{W_1} = \frac{nRT_2 \ln(V_f/V_i)}{nRT_1 \ln(V_f/V_i)} = \frac{T_2}{T_1} = \frac{450\text{ K}}{300\text{ K}} = 1.5

This confirms that (A) 1.5 is correct. (B) 2.0 represents the volume ratio

V_f/V_i

, which is a trap for students who confuse the given volume ratio with the work ratio. (C) 3.0 might result from incorrectly adding the temperature ratio to something else, or from computational errors. (D) 0.67 is the reciprocal of the correct answer (

1/1.5

), which occurs if you accidentally invert the temperature ratio. Study tip: For isothermal processes, work is directly proportional to temperature when all other factors (amount of gas, volume ratios) are identical. Always check that you're using the correct formula and putting the higher temperature in the numerator when asked for a ratio of "Process 2 to Process 1" where Process 2 occurs at higher temperature.

Question 18

During an isothermal compression of an ideal gas, the internal energy change is zero while heat is transferred out of the system. If the gas is compressed from 5.0 L to 1.0 L at constant temperature 300 K, and 2500 J of work is done on the gas, what can be concluded about the heat transfer?

Heat absorbed by the gas is 2500 J since work and heat must balance
Heat released by the gas is 2500 J according to the first law constraint (correct answer)
Heat transfer is zero because the process is isothermal and reversible
Heat released is less than 2500 J due to internal energy storage effects

Explanation: From the first law:

\Delta U = Q - W

. For an isothermal process with an ideal gas,

\Delta U = 0

since internal energy depends only on temperature. Therefore:

0 = Q - W

, so

Q = W

. Since 2500 J of work is done ON the gas,

W = -2500

J (negative because work is done on the system). Thus

Q = -2500

J, meaning 2500 J of heat is released by the gas. Choice A has the wrong sign. Choice C ignores the first law requirement. Choice D incorrectly assumes internal energy changes.

Question 19

An ideal gas undergoes a cyclic process consisting of two isothermal steps at different temperatures connected by two isochoric steps. During the isothermal expansion at 400 K, the gas does 1200 J of work. During the isothermal compression at 300 K, 800 J of work is done on the gas. What is the net work done by the gas in one complete cycle?

Net work is 400 J since isochoric processes contribute no work to the total (correct answer)
Net work is 2000 J representing the sum of all work transfers in the cycle
Net work is 1200 J because only expansion processes contribute positive work output
Net work is zero because the gas returns to its initial state after the cycle

Explanation: In a complete thermodynamic cycle, the net work is the algebraic sum of work done in each process. Isothermal expansion:

W_1 = +1200

J (work done by gas). Isothermal compression:

W_2 = -800

J (work done on gas). Isochoric processes:

W = 0

since

dV = 0

. Net work =

1200 + (-800) + 0 + 0 = 400

J. Choice B incorrectly adds absolute values. Choice C ignores the compression work. Choice D confuses the return to initial state (which makes

\Delta U = 0

) with zero net work.

Question 20

An ideal gas sample undergoes an isothermal expansion at 400 K, during which it absorbs 2000 J of heat from the surroundings. Subsequently, the gas undergoes an isothermal compression at the same temperature back to its original volume, releasing 1800 J of heat. What can be concluded about these processes?

The expansion was irreversible because heat absorption exceeded the theoretical maximum
Both processes were reversible since they occurred at constant temperature throughout
The compression was irreversible since less heat was released than absorbed during expansion (correct answer)
The net effect violates conservation of energy since heat values don't match exactly

Explanation: For a complete cycle returning to the initial state,

\Delta U_{total} = 0

. The net work done by the gas equals the net heat absorbed:

W_{net} = Q_{net} = 2000 - 1800 = 200

J. In a reversible isothermal cycle, the work done during expansion should equal the work required for compression back to the original state, making

Q_{expansion} = Q_{compression}

. Since they differ, the compression process was irreversible (required extra work due to irreversibilities). Choice B ignores the heat difference. Choice C incorrectly blames the expansion. Choice D misunderstands that energy is still conserved.