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AP Physics 2 : AP Physics 2

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Example Questions

Example Question #41 : Thermodynamics

You are at the top of Mt. Whitney holding onto a balloon of volume 1.12 L . The temperature is $19^{\circ}C$ , and the air pressure is .83 atm . You then drive to Death Valley, where the temperature is $35 ^{\circ}C$ , and the air pressure is 1.11 atm . What is the new volume of the balloon?

Possible Answers:

1.62 L

.43 L

None of these

.83 L

.88 L

Correct answer:

.88 L

Explanation:

We will use the combined gas law:

$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$

With "1" referring to Mt Whitney, and "2" referring to Death Valley, we can rearrange the equation to get

$\frac{P_1 V_1 T_2}{T_1 P_2}=V_2$

We need to convert both temperatures to Kelvin

$19^{\circ}C+273=292^{\circ}K$

$35^{\circ}C+273=308^{\circ}K$

We plug in our values.

$\frac{.83 \cdot1.12\cdot 308}{292\cdot 1.11}=V_2$

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Example Question #1 : Ideal Gas Law

A balloon is in a room that has a constant pressure of 100 kPa and constant temperature of $20^{\circ}C$ . How many moles of air must be put into the balloon for 20kJ of work to be done on the balloon?

Possible Answers:

9.1 mol

4.4 mol

8.2 mol

12.1mol

6.7 mol

Correct answer:

8.2 mol

Explanation:

We know that pressure and temperature are constant. Therefore, we can use the following formula for work:

$W=P\Delta V$

Write out the ideal gas law:

$P\Delta V=\Delta nRT$

Rearrange for moles:

$\Delta n = \frac{P\Delta V}{RT}$

Substitute in our expression for work:

$\Delta n = \frac{W}{RT}$

Now, plug in values for each variable:

$\Delta n = \frac{20kJ}{(8.314\frac{J}{mol\cdot K})(293.15K)} = 8.2 mol$

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Example Question #1 : Ideal Gas Law

Suppose that a gas originally at standard temperature and pressure undergoes a change in which its pressure is quadrupled while its temperature is cut in half. What change in volume does the gas experience during this process?

Possible Answers:

Decrease by a factor of 2

The volume of the gas will not change

Increase by a factor of 8

Decrease by a factor of 8

Increase by a factor of 2

Correct answer:

Decrease by a factor of 8

Explanation:

To solve this problem, we'll need to use the ideal gas equation:

PV=nRT

We are told that the gas undergoes a change in which its pressure quadruples and its temperature halves. Therefore:

$P^{'}=4P$

and

$T^{'}=\frac{1}{2}T$

Furthermore, we can set the ideal gas equation to solve for volume:

$V=\frac{nRT}{P}$

and

$V^{'}=\frac{nRT^{'}}{P^{'}}$

If we plug in the values from above, we obtain:

$V^{'}=\frac{nR(\frac{1}{2}T)}{4P}=\frac{nRT}{8P}=\frac{V}{8}$

Therefore, we can see that the new volume is $\frac{1}{8}$ of its original value.

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Example Question #3 : Ideal Gas Law

An ideal gas is kept in a container at $50^{\circ}C$ and 796Torr . How many moles of the gas are in the container?

760Torr=1atm

Possible Answers:

79.60mol

There is not enough information to determine the number of moles

0.001949mol

969.5mol

0.197mol

Correct answer:

0.197mol

Explanation:

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

PV=nRT

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.

796Torr=1.047atm

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

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

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

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

0.197mol&=n

Therefore, there are 0.197mol of gas in the container.

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Example Question #2 : Ideal Gas Law

At $40^{\circ}C$ , the volume of a gas is 300mL . The temperature of the gas gets raised to $55^{\circ}C$ with no change in pressure. What is the new volume of the gas?

Possible Answers:

412.5 mL

515.3 mL

286.2 mL

314.0 mL

There is not enough information to determine the new volume

Correct answer:

314.0 mL

Explanation:

When the only properties of an ideal gas that are changing are volume and temperature, we use Charles' Law (a derivative of the Ideal Gas Law). Charles' Law is as follows:

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

We're given all but the new volume, V_2 . To find the new volume, we rearrange the equation.

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

$\\ \frac{0.300L\cdot (55^oC+273.15)}{(40^oC+273.15)}&=V_2$

The addition of 273.15 is to convert the Celsius units to Kelvin.

0.314L&=V_2

314.0mL &=V_2

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Example Question #5 : Ideal Gas Law

The pressure of a sample of gas is 5atm in a 1.2L sealed, flexible container. If the pressure gets raised to 8atm at a constant temperature, what is the new volume?

Possible Answers:

1.4L

1.0L

There is no volume change

0.75 L

1.92L

Correct answer:

0.75 L

Explanation:

Because the only properties of the gas that are changing are the pressure and volume¹, we use Boyle's Law, a derivative of the Ideal Gas Law. Boyle's Law states

P_1V_1 = P_2V_2

Since can be in atmospheres, and can be in Liters, we don't have to convert any units. Instead, we just rearrange the equation to solve for V_2 , and plug in our numbers.

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

$\frac{5atm \cdot 1.2L}{8atm}&=V_2$

0.75L&=V_2

The new volume is 0.75L .

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Example Question #1 : Ideal Gas Law

An airship has a volume of 300000m^3 . How many kilograms of hydrogen would fit in it at 0.95atm and $25^{\text o}C$ ?

Possible Answers:

116000kg

116153.32kg

23484.21kg

233000kg

None of these

Correct answer:

23484.21kg

Explanation:

Use the ideal gas equation:

PV=nRT

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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Example Question #1 : Ideal Gas Law

Assuming ideal gas behavior, find the density of pure oxygen gas at 1.33atm and $-10^{\text o}C$ .

Possible Answers:

None of these

$1.97\frac{g}{L}$

$0.98\frac{g}{ml}$

$1.97\frac{kg}{L}$

$0.98\frac{g}{L}$

Correct answer:

$1.97\frac{g}{L}$

Explanation:

For simplicity, we will assume 1mol .

Rearrange the ideal gas equation to solve for volume.

PV=nRT

$\frac{nRT}{P}=V$

Plug in known values and solve.

V=16.2 L

Use volume to solve for density.

1 mol O_2 = 32 g

$\frac{32g}{16.2L}=1.97\frac{g}{L}$

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Example Question #1 : Ideal Gas Law

Assuming ideal gas behavior, determine the volume of 37g of methane gas at $15^{\text o}C$ and 0.91atm .

Possible Answers:

30.1L

None of these

59.0L

15.0L

Correct answer:

59.0L

Explanation:

Use the ideal gas equation and rearrange, solving for volume.

PV=nRT

$V=\frac{nRT}{P}$

Find by using the molar mass of methane.

$37g\: CH_4\cdot\frac{1mol}{16g\:CH_4}=2.31mol$

Plug in known values into the rearranged ideal gas equation and solve.

$V=\frac{2.31mol\cdot0.08206\frac{L\cdot atm}{mol\cdot K}\cdot 283K}{0.91atm}$

$V\approx59.0L$

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Example Question #9 : Ideal Gas Law

In a room where the temperature is T=300K , a football has been inflated to a gauge pressure of P=89kPa . The football is then taken to the field, where the temperature is T=270K . What will the football's gauge pressure be when its temperature becomes equal to the temperature of the air on the field? Assume the air follows the ideal gas law and that the atmospheric pressure that day was P=101kPa .

Possible Answers:

P=110kPa

P=89.1kPa

P=98.9kPa

P=40 kPa

P=70kPa

Correct answer:

P=70kPa

Explanation:

Start by converting to Pascals:

$P_F=89\:kPa* \left(\frac{1000Pa}{kPa}\right)=89000Pa$

For the atmosphere: P_A=101000 Pa

Find the absolute pressure in the football:

P=P_A+P_F=101000Pa+89000Pa=190000Pa

Write the ideal gas law for the football in the locker room:

P_1V_1=nRT_1

Solve for , the constants that won't change as the air cools:

$nR=\frac{P_1V_1}{T_1}$

Write the ideal gas law for the football on the field:

P_2V_2=nRT_2

Substitute from before:

$P_2V_2=\frac{P_1V_1}{T_1}T_2$

Recognize that the volume does not change, so those terms cancel:

$P_2=P_1\left(\frac{T_2}{T_1}\right)$

$P_2=190000Pa(\frac{270K}{300K})=171000Pa$

Convert from absolute pressure to gauge pressure:

P_F=P_2-1atm=171000Pa-101000Pa=70000Pa=70kPa

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All AP Physics 2 Resources

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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #41 : Thermodynamics

Example Question #1 : Ideal Gas Law

Example Question #1 : Ideal Gas Law

Example Question #3 : Ideal Gas Law

Example Question #2 : Ideal Gas Law

Example Question #5 : Ideal Gas Law

Example Question #1 : Ideal Gas Law

Example Question #1 : Ideal Gas Law

Example Question #1 : Ideal Gas Law

Example Question #9 : Ideal Gas Law

All AP Physics 2 Resources

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