AP Physics 2 - Properties of Ideal Gases

Determine the root mean square velocity of chlorine gas at $150^\circ C$ .

$k=1.38*10^{-23}\frac{J}{^\circ K}$

$385\frac{m}{s}$

None of these

$717\frac{m}{s}$

$844\frac{m}{s}$

$138\frac{m}{s}$

Explanation

Using,

$v_{rms}=\sqrt{\frac{3kT}{m}}$

Where

is the Boltzmann constant, $1.38*10^{-23}\frac{J}{^\circ K}$

is the temperature, in Kelvin

is the mass of a single molecule, in kilograms

Finding mass of a single chlorine molecule:

$\frac{70.9grams}{mole}*\frac{1 mole}{6.02*10^{23} molecules}*\frac{1kg}{1000g}=1.18*10^{-25}\frac{kg}{molecule}$

Plugging in values:

$v_{rms}=\sqrt{\frac{3*1.38*10^{-23}*423}{1.18*10^{-25}}}$

$v_{rms}=385\frac{m}{s}$

A balloon is filled with nitrogen at a temperature of $15\ ^{\circ}C$ to a pressure of $1\textup{ atm}$ . If the temperature is increased to $25\ ^{\circ}C$ and the pressure is increased to $1.5\textup{ atm}$ , by what factor does the density of the nitrogen increase by?

Explanation

We will start with the ideal gas law for this problem:

Since the problem statement is asking us for a change in density, we will need to manipulate the ideal gas law to get this unit. We will start by multiplying both sides of the expression by the molecular weight of nitrogen:

Now rearranging we get:

$\frac{m}{V}=\rho = \frac{P(MW)}{RT}$

We can apply this to both scenarios, with density, pressure, and temperature being the only changing variables:

$\rho_1 = \frac{P_1(MW)}{RT_1}$

$\rho_2 = \frac{P_2(MW)}{RT_2}$

Dividing the second expression by the first and eliminating common variables, we get:

$\frac{\rho_2}{\rho_1} = \frac{\frac{P_2}{t_2}}{\frac{P_1}{T_1}} = \frac{P_2T_1}{P_1T_2}$

Plugging in our values and making sure our temperatures are in Kelvin, we get:

$\frac{\rho_2}{\rho_1} = \frac{(1.5atm)(288C)}{(1atm)(298C)}$

$\frac{\rho_2}{\rho_1} = 1.45$

We have of helium gas at . We decrease the pressure to , while keeping the temperature the same. What is the new volume?

Explanation

We will use the relationship:

Rearrange to solve for the final volume:

$\frac{P_1V_1}{P_2}=V_2$

Plug in our values and solve.

$\frac{1.3atm*13L}{.85atm}=V_2$

A real gas becomes more like an ideal gas at __________ temperatures and __________ pressures.

higher . . . lower

lower . . . higher

lower . . . lower

higher . . . higher

Real gases can never be made to act more like ideal gases

Explanation

When the temperature is higher, the kinetic energy of particles is higher. Because their energy is higher, they bounce more energetically, letting them ignore most intermolecular forces (which is something ideal gases don't experience).

An important assumption for ideal gases is that the volume of the particles is negligible compared to the volume of their container. For real gases, as pressure increases, the particles get closer together, and their volume gets less negligible when compared to the volume of the container. This is why real gases behave more ideally at lower pressures.

In an enclosed space capsule, the temperature increases from $13^\circ C$ to $27^\circ C$ . Determine the ratio of the final to initial pressures: $\frac{P_{final}}{P_{initial}}$ .

Explanation

Convert to

Use Gay-Lussac's law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Plug in values:

$\frac{P_1}{286}=\frac{P_2}{300}$

$\frac{P_2}{P_1}=1.05$

Which of the following is the correct formula to find the internal energy of an ideal gas?

$U=\frac{3}{2}NkT$

$U=\frac{1}{2}NkT$

$U=\frac{3}{2}kT$

$U=\frac{1}{2}kT^2$

$U=\sqrt{\frac{3kT}{M}}$

Explanation

An ideal gas has no molecular interactions besides perfectly elastic collisions. The energy is the combined kinetic energies of ideal gas molecules since there is no potential energy. For point molecules of the monatomic gas, the equation can be written as:

$U=\frac{3}{2}NkT$

Note that $K_{{avg}}=\frac{3}{2}kT$ is the equation for average kinetic energy also in Joules.

What is the temperature in Kelvin for 1mol of gas at 5atm and a volume of 10L?

$1{ atm}= 101.3{ kPa}= 101325 \frac{{N}}{{m}^2}$

$R= 8.314{ }\frac{{L}\cdot {kPa}}{{mol} \cdot {K}}$

Explanation

Use the ideal gas law equation.

Manipulate the equation, substitute the givens, and solve for temperature. Write the gas constants necessary to solve the problem.

$T=\frac{PV}{nR}$

$1{ atm}= 101.3{ kPa}= 101325 \frac{{N}}{{m}^2}$

$R= 8.314{ }\frac{{L}\cdot {kPa}}{{mol} \cdot {K}}$

$T=\frac{PV}{nR}= \frac{(5{ atm})(\frac{101.3{ kPa}}{1{ atm}})(10{ L})}{(1{ mol})(8.314{ }\frac{{L}\cdot {kPa}}{{mol} \cdot {K}})}$

Determine the average velocity of radon gas atoms at $35\ ^\circ \textup{C}$

$\textup{R}=8.314\ \frac{\textup{J}}{\textup{mol}\cdot \textup{K}}$

$186.3\ \frac{\textup{m}}{\textup{s}}$

$219.4\ \frac{\textup{m}}{\textup{s}}$

$345.6\ \frac{\textup{m}}{\textup{s}}$

$398.5\ \frac{\textup{m}}{\textup{s}}$

$444.9\ \frac{\textup{m}}{\textup{s}}$

Explanation

Using the equation for the root-mean-square speed:

$v=\sqrt{\frac{3RT}{M}}$

$\textup{R}=8.314\ \frac{\textup{J}}{\textup{mol}\cdot \textup{K}}$

is the temperature in Kelvin

is the molar mass of the gas, in kilograms per mole

Converting Celsius to Kelvin and plugging in values:

$v=\sqrt{\frac{3*8.314*308}{.222}}$

$v=186.3\frac{m}{s}$

We have of gas at $10 ^\circ C$ . We increase the temperature to $30 ^\circ C$ , while keeping the pressure constant. What is the new volume?

Explanation

We will use the relationship:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Rearrange to solve for the final volume:

$\frac{V_1 T_2}{T_1}={V_2}$

Convert our temperatures to kelvin:

$10^{\circ}C+273=283^{\circ}K$

$30^{\circ}C+273=303^{\circ}K$

Plug in our values and solve.

$\frac{15L* 303^{\circ}K}{283^{\circ}K}={V_2}$

Determine the average velocity of chlorine molecules at $35^\circ C$ .

$329\frac{m}{s}$

None of these

$277\frac{m}{s}$

$93.1\frac{m}{s}$

$151\frac{m}{s}$

Explanation

Using

$v=\sqrt{\frac{3RT}{M_m}}$

is the temperature in Kelvin

is the gas constant

is the molar mass in kilograms

Converting to Kelvin and plugging in values:

$v=\sqrt{\frac{3*8.314*308}{.0709}}$

$v=329\frac{m}{s}$

Properties of Ideal Gases

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