Ideal Gas Law - AP Physics 2

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Question

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?

Answer

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

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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Question

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

Answer

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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Question

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

Answer

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.

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Question

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

Answer

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, . 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.

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Question

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

Answer

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

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 , and plug in our numbers.

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

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

The new volume is .

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Question

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

Answer

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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Question

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

Answer

For simplicity, we will assume .

Rearrange the ideal gas equation to solve for volume.

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

Plug in known values and solve.

Use volume to solve for density.

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

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Question

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

Answer

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

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

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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Question

A balloon filled is filled with pure nitrogen gas. The balloon is determined to have a volume of on a day when the temperature is $20^{\circ}C$ , and the air pressure is .

How many nitrogen molecules are present?

Answer

We will use our ideal gas equation.

Where is the pressure

is the volume

is the number of moles

is the gas constant

is the temperature in Kelvin

We rearrange the equation to solve for n

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

A common mistake is using the wrong gas constant, . We need to use

$.0821 \frac{L\cdot atm}{K\cdot mol}$

We convert our temperture from Celsius to Kelvin

$20^{\circ}C+273=293^{\circ}K$

We plug in our values

$n=\frac{.85.75}{.0821293}$

We then need to multiply by Avogadros number to convert to number of molecules.

$.0265moles\cdot \frac{6.02\times10^{23}molecules}{mole}=n$

$1.57 \times 10^{22} molecules=n$

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Question

How many moles of nitrogen gas are in a tank with a pressure of at ?

$R=0.0821\frac{L\cdot atm}{mol\cdot K}$

Answer

We will use the ideal gas equation:

Where is the pressure in atm, is the volume in liters, is the number of moles of gas, $R=0.0821\frac{L\cdot atm}{mol\cdot K}$ , and is the temperature in Kelvin.

Rearrange the equation to solve for moles:

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

Plug in known values and solve.

$n=\frac{510}{0.0821302}$

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Question

What is the volume of of nitrogen gas (diatomic) at $24^{\circ}C$ and ?

$R=.0821 \frac{(atmL)}{(Kmol)}$

Answer

We will use the equation:

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atmL)}{(Kmol)}$

is the temperature in Kelvin.

We need to convert the temperature to Kelvin.

$24^{\circ}C+273=298^{\circ}K$

Rearrange our original equation for volume:

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

We need to find the moles of nitrogen gas. We divide the mass by the molar mass of nitrogen, which is $28 \frac{g}{mol}$ .

$\frac{(1000 g)}{(28 g/mol)}=35.7 moles$

Plug in our values to our original equation and solve.

$V=\frac{35.7.0821297}{.98}$

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Question

What is the volume of of gaseous water at $125^{\circ}C$ and ?

$R=.0821 \frac{(atmL)}{(Kmol)}$

Answer

We will use the equation

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atmL)}{(Kmol)}$

is the temperature in Kelvin

We need to convert the temperature to Kelvin

Rearrange our original equation for volume:

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

We need to find the moles of nitrogen gas. We divide the mass by the molar mass of water, which is $18 \frac{g}{mol}$ .

$\frac{1000 g}{18 \frac{g}{mol}}=55.6mol$

Plug in our values to our rearranged original equation and solve.

$\frac{55.60821398}{1.02}=V$

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Question

What is the mass of of helium gas at $21^\circ C$ and ?

$R=.0821 \frac{(atmL)}{(Kmol)}$

Answer

We will use the equation:

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atmL)}{(Kmol)}$

is the temperature in Kelvin

We need to convert the temperature to Kelvin.

$21^{\circ}C+273=294^{\circ}K$
Rearrange to solve for moles:

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

Plug in our values and solve

$\frac{1.0316.5}{.0821294}=n$

Then, we need to multiply by the molar mass of helium; recall that helium is not a diatomic gas.

$.704 moles He * \frac{4g He}{mol} = 2.81 g$

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Question

What is the mass of of oxygen gas at $-10^{\circ}C$ and ?

$R=.0821 \frac{(atmL)}{(Kmol)}$

Answer

We will use the equation:

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atmL)}{(Kmol)}$

is the temperature in Kelvin

Rearrange to solve for moles:

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

Plug in our values and solve

$\frac{.956}{.0821263}=n$

Then, we need to multiply by the molar mass of oxygen gas, which is diatomic.

$.263 moles O_2 * \frac{32g O_2}{mol} = 8.45 g$

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Question

What is the density of hydrogen gas at $-10 ^{\circ}C$ and ?

$R=.0821 \frac{(atmL)}{(Kmol)}$

Answer

We will use the equation

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atmL)}{(Kmol)}$

is the temperature in Kelvin

We need to convert celsius to Kelvin

$-10^{\circ}C+273=263^{\circ}K$

Then, we will assume to make our calculations easier.

Rearrange for volume:

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

Plug in our values and solve

$\frac{1.0821263}{.95}=V$

Again, assume and multiply by the molar mass of hydrogen gas, $\frac{2g}{mol}$ :

$1mol*\frac{2gH_2}{1molH_2}=2 g$

We divide mass by volume to get density.

$D=\frac{2g}{22.7L}$

$D=88\frac{mg}{L}$

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Question

What is the density of neon gas at $-5^{\circ} C$ and ?

$R=.0821 \frac{(atmL)}{(Kmol)}$

Answer

We will use the equation:

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atmL)}{(Kmol)}$

is the temperature in Kelvin

First we need to convert our temperature to Kelvin:

We will assume to make our calculations easier.

Rearrange for volume:

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

Plug in our values and solve.

$\frac{1mol.0821 \frac{(atmL)}{(Kmol)}268^{\circ}K}{.97atm}=V$

Again, assume and multiply by the molar mass of neon $\frac{20.2g}{mol}$ :

$1molNe\frac{20.2gNe}{molNe}=20.2 gNe$

We divide mass by volume to find density.

$D=\frac{M}{V}$

$D=\frac{20.2g}{22.7L}$

$.885\frac{g}{L}=D$

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Question

In a room where the temperature is , a football has been inflated to a gauge pressure of . The football is then taken to the field, where the temperature is . 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 .

Answer

Start by converting to Pascals:

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

For the atmosphere:

Find the absolute pressure in the football:

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

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:

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:

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Question

Which of the following graphs shows an incorrect relationship for an ideal gas?

Answer

This question is asking for us to identify which graph shows an incorrect relationship between volume, pressure, and temperature of an ideal gas. To distinguish between which graphs are correct, and which one isn't, we'll need to use the ideal gas law.

It's also important to note that each graph in the answer choices shows a linear, positive relationship between the x and y-variables.

The only graph that shows an incorrect relationship is the graph that has as the y-axis and $\frac{V}{T}$ as the x-axis. From the ideal gas law shown above, we would expect a positive linear relationship of a graph showing vs. $\frac{T}{V}$ . If we instead have vs. $\frac{V}{T}$ , the graph would have a negative relationship.

All of the other graphs shown in the remaining answer choices show correct relationships. vs. $\frac{T}{V}$ gives a positive relationship. Likewise, vs. gives a positive relationship. And finally, vs. $\frac{T}{P}$ also gives a positive relationship.

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Question

How many moles of gas are there in a container at a temperature of at atmospheric pressure?

Answer

In this question, let's start with what we know and what we want to know. We're given the volume of the gas in a container, as well as the pressure and the temperature. Then, we're asked to find the number of moles of gas that are in this container.

To answer this question, we'll need to make use of the ideal gas equation.

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

Plugging in the values that we know, we can calculate our answer.

$n=\frac{(1: atm)(12: L)}{(0.08206: \frac{L\cdot atm}{mol\cdot K})(300: K)}$

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Question

of gas are heated from $-2^\circ C$ to $25^\circ C$ and the pressure increases from to . Determine the volume in the final state.

$R=.0821\frac{Latm}{moleK}$

Answer

Only the final state data will be needed Convert Celsius to Kelvin:

Use the ideal gas law:

Where

is the pressure in

is the volume in

is the number of moles

is the gas constant, $.0821\frac{Latm}{moleK}$

is the temperature in

Plug in values and solve:

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