Thermodynamics

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AP Physics 2 › Thermodynamics

Questions 1 - 10

In a $15\ \textup{L}$ rigid, sealed container, there is 1 mole of an ideal gas. The container is initially at $25^\circ \textup{C}$ . If the gas is heated to $50^\circ\textup{C}$ , what is the new pressure in the container?

$1.768 \textup{ atm}$

$1.431 \textup{ atm}$

$1.631 \textup{ atm}$

$2.451 \textup{ atm}$

$1.122 \textup{ atm}$

Explanation

To determine the answer, we must do several steps with the ideal gas law:

For this problem, the pressure and temperature are the only things to change. Therefore, we can rearrange the equation to account for this.

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

However, we don't know what the value of P1 is just yet. To determine it, we must use the values we do know to solve for the initial pressure.

$\begin{align*} P&=\frac{nRT}{V} \ &=\frac{1*0.0821*298}{15} \ &= 1.631\ atm \end{align*}$

Now, we have P1. Using this, along with T1 and T2, we can solve for P2.

$\begin{align*} P_2&=\frac{T_2*P_1}{T_1} \ &= \frac{323*1.631}{298}\ &= 1.768\ atm \end{align*}$

$1.768 \textup{ atm}$

$1.431 \textup{ atm}$

$1.631 \textup{ atm}$

$2.451 \textup{ atm}$

$1.122 \textup{ atm}$

Explanation

To determine the answer, we must do several steps with the ideal gas law:

For this problem, the pressure and temperature are the only things to change. Therefore, we can rearrange the equation to account for this.

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

However, we don't know what the value of P1 is just yet. To determine it, we must use the values we do know to solve for the initial pressure.

$\begin{align*} P&=\frac{nRT}{V} \ &=\frac{1*0.0821*298}{15} \ &= 1.631\ atm \end{align*}$

Now, we have P1. Using this, along with T1 and T2, we can solve for P2.

$\begin{align*} P_2&=\frac{T_2*P_1}{T_1} \ &= \frac{323*1.631}{298}\ &= 1.768\ atm \end{align*}$

Which of the following best defines the zeroth law of thermodynamics in variable form?

$\textup{X}=\textup{Y},\textup{Y}=\textup{Z},\textup{X}=\textup{Z}$

$\Delta U = \textup{Q}-\textup{W}$

$\textup{H}=\textup{U}+\textup{PV}$

$\eta_{th}<100%$

$\Delta S= \frac{\Delta Q}{t}$

Explanation

The zeroth law of thermodynamics states that if an object X is in thermal equilibrium with object Y, and object Y also in thermal equilibrium with object Z, then object X must also be in thermal equilibrium with object Z.

$\Delta U = \textup{Q}-\textup{W}$ refers to the first law of thermodynamics.

$\eta_{th}<100%$ refers to the second law of thermodynamics.

$\textup{H}=\textup{U}+\textup{PV}$ is the enthalpy equation.

$\Delta S= \frac{\Delta Q}{t}$ refers to entropy.

The best choice is:

$\textup{X=Y, Y=Z, X=Z.}$

Which of the following best defines the zeroth law of thermodynamics in variable form?

$\textup{X}=\textup{Y},\textup{Y}=\textup{Z},\textup{X}=\textup{Z}$

$\Delta U = \textup{Q}-\textup{W}$

$\textup{H}=\textup{U}+\textup{PV}$

$\eta_{th}<100%$

$\Delta S= \frac{\Delta Q}{t}$

Explanation

$\Delta U = \textup{Q}-\textup{W}$ refers to the first law of thermodynamics.

$\eta_{th}<100%$ refers to the second law of thermodynamics.

$\textup{H}=\textup{U}+\textup{PV}$ is the enthalpy equation.

$\Delta S= \frac{\Delta Q}{t}$ refers to entropy.

The best choice is:

$\textup{X=Y, Y=Z, X=Z.}$

An ideal gas is kept in a container at $50^{\circ}C$ and . How many moles of the gas are in the container?

There is not enough information to determine the number of moles

Explanation

Because this is an ideal gas, we can use the Ideal Gas Law to determine its state.

The value for is sometimes tricky to determine, because it has several values depending on the units being used. The two main values for that are used are:

$8.314\frac{J}{mol\cdot K}$ and $\ 0.0821 \frac{L\cdot atm}{K\cdot mol}$

Because we have units in Liters, and we can convert our temperature and pressure to Kelvin and atmospheres respectively, we use the second value of .

First, let's convert our values to usable units.

$K= {^{\circ}}C+273.15\rightarrow K=323.15$

Because we're trying to find moles of gas, we can rearrange the ideal gas equation to equal moles, and plug in our values.

$\frac{PV}{RT}&=n$

$\frac{1.047atm\cdot 5L}{0.0821\frac{L\cdot atm}{K\cdot mol}\cdot 323.15K}&=n$

Therefore, there are of gas in the container.

An ideal gas is kept in a container at $50^{\circ}C$ and . How many moles of the gas are in the container?

There is not enough information to determine the number of moles

Explanation

Because this is an ideal gas, we can use the Ideal Gas Law to determine its state.

The value for is sometimes tricky to determine, because it has several values depending on the units being used. The two main values for that are used are:

$8.314\frac{J}{mol\cdot K}$ and $\ 0.0821 \frac{L\cdot atm}{K\cdot mol}$

Because we have units in Liters, and we can convert our temperature and pressure to Kelvin and atmospheres respectively, we use the second value of .

First, let's convert our values to usable units.

$K= {^{\circ}}C+273.15\rightarrow K=323.15$

Because we're trying to find moles of gas, we can rearrange the ideal gas equation to equal moles, and plug in our values.

$\frac{PV}{RT}&=n$

$\frac{1.047atm\cdot 5L}{0.0821\frac{L\cdot atm}{K\cdot mol}\cdot 323.15K}&=n$

Therefore, there are of gas in the container.

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:

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

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

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:

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

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

An airship has a volume of . How many kilograms of hydrogen would fit in it at and $25^{\text o}C$ ?

None of these

Explanation

Use the ideal gas equation:

Convert the volume into liters in order to use our ideal gas constant: $R=0.0821 \frac{L\cdot atm}{K\cdot mol}$

$300000m^3 \cdot \frac{100^3 cm^3}{1m^3}\cdot\frac{1ml}{1cm^3}\cdot\frac{1 L}{1000mL}=3.0\cdot10^8 L$

Rearrange the ideal gas equation to solve for , then plug in known values and solve.

$\frac{0.95atm\cdot 3.0\cdot10^8 L}{298K\cdot 0.0821 L\cdot atm \cdot K^{-1} \cdot mol^{-1}}=n$

$n\approx11648914 mol$

$11648914mol:H_{2}_{(g)} \cdot\frac{2.016g}{1mol}\approx 23484210g$

$23484210g\cdot\frac{1kg}{1000g}=23484.21kg$

An airship has a volume of . How many kilograms of hydrogen would fit in it at and $25^{\text o}C$ ?

None of these

Explanation

Use the ideal gas equation:

Convert the volume into liters in order to use our ideal gas constant: $R=0.0821 \frac{L\cdot atm}{K\cdot mol}$

$300000m^3 \cdot \frac{100^3 cm^3}{1m^3}\cdot\frac{1ml}{1cm^3}\cdot\frac{1 L}{1000mL}=3.0\cdot10^8 L$

Rearrange the ideal gas equation to solve for , then plug in known values and solve.

$\frac{0.95atm\cdot 3.0\cdot10^8 L}{298K\cdot 0.0821 L\cdot atm \cdot K^{-1} \cdot mol^{-1}}=n$

$n\approx11648914 mol$

$11648914mol:H_{2}_{(g)} \cdot\frac{2.016g}{1mol}\approx 23484210g$

$23484210g\cdot\frac{1kg}{1000g}=23484.21kg$

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