Ideal Gas Law

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AP Physics 2 › Ideal Gas Law

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.

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

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 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.

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

$n\approx11648914 mol$

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

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

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 .

Explanation

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:

By what factor does the volume of an ideal gas change if its temperature increases by 50% and pressure quadruples?

Explanation

We will use the ideal gas law to solve this problem:

We are asked to find the change in volume, so let's rearrange for that:

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

Then applying this to the initial and final scenarios:

$V_1 = \frac{nRT_1}{P_1}$

$V_2 = \frac{nRT_2}{P_2}$

Then taking the ratio of scenario 2 to scenario 1:

$\frac{V_2}{V_1}= \frac{\frac{nrT_2}{P_2}}{\frac{nrT_1}{P_1}} = \frac{T_2P_1}{T_1P_2}$

Where:

Substituting these in, we get:

$\frac{V_2}{V_1}=\frac{1.5T_1P_1}{T_14P_1} = \frac{1.5}{4} = 0.375$

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.

$R=.0821\frac{L*atm}{mole*K}$

Explanation

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, $.0821\frac{L*atm}{mole*K}$

is the temperature in

Plug in values and solve:

A mixture of gas has a volume of $40\textup{ L}$ , a pressure of $2.3\textup{ atm}$ , and is at a temperature of $40\ ^{\circ}C$ . If the gas mixture is 80% nitrogen and 20% oxygen, how many moles of nitrogen are there?

$2.9\textup{ mol}$

$3.6\textup{ mol}$

$2.1\textup{ mol}$

$1.4\textup{ mol}$

$0.81\textup{ mol}$

Explanation

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

Then rearranging for total moles:

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

$n = \frac{(2.3atm)(40L)}{\left ( 8.314\frac{J}{k\cdot mol}\right )(313K)}\left ( \frac{101325Pa}{atm}\right )\left ( \frac{1m^3}{1000L}\right )$

Then multiplying by 80% to get the number of moles of nitrogen:

$n_{nitrogen}=0.80n = 0.80(3.58)$

$n_{nitrogen} = 2.9 mol$

How many oxygen molecules are there in a $5\textup{L}$ tank at $16\ ^\circ\textup{C}$ and $200\textup{ atm}$ ?

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

$2.54*10^{25}\textup{ molecules}$

None of these

$3.10*10^{25}\textup{ molecules}$

$3.40*10^{25}\textup{ molecules}$

$1.52*10^{25}\textup{ molecules}$

Explanation

Using ideal gas law:

Converting Celsius to Kelvin and plugging in values:

$42moles*\frac{6.02*10^{23}molecules}{1 mole}=2.54*10^{25}molecules$

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 .