AP Chemistry - Ideal Gas Law

If the pressure of a sample of one mole of an ideal gas is increased from 2atm to 3atm at a constant volume, and the initial temperature was 20˚C, what is the final temperature of the sample?

439.5K

30K

195.3K

303K

Explanation

Because the mass and volume of the sample of the ideal gas are kept constant, a change in pressure causes only a direct change in the temperature. This can be derived from the following ideal gas equation:.

$\frac{T_{2}}{T_{1}}=\frac{P_{2}}{P_{1}}$

$\frac{T_{final}}{293 K}=\frac{3 atm}{2 atm}$

$T_{final}=439.5 K$

If a container holds 1mol of hydrogen, 2.5mol of helium, and 2mol of oxygen at a total pressure of 4atm, what is the partial pressure of the oxygen gas?

Explanation

According to Dalton's law of partial pressures, when two or more gases are in one container without chemical interaction, each behaves independently of the others. The partial pressure of oxygen can, therefore, be found by multiplying the molar fraction of oxygen in the container by the total pressure of the three gases.

$X_{oxygen}= \frac{2mol\ O_2}{1mol\ H_2+2.5mol\ He+2mol\ O_2}= 0.37$

$X_{oxygen}*P=P_{oxygen}$

$0.37 * 4 atm\approx 1.45 atm$

Ammonia is created according to the balanced equation below.

$3H_{2} + N_{2} \rightarrow 2NH_{3}$

The reaction is allowed to take place in a rigid container. Eight moles of hydrogen gas are mixed with two moles of nitrogen gas. The initial pressure exerted on the container is 5atm.

Assuming the reaction runs to completion, what will the pressure exerted on the vessel be after the reaction takes place?

Explanation

Since the total pressure is dependent on the number of moles in the container, we can use the ratio of moles before and after the reaction to determine the final pressure in the container. There are initially ten moles of gas in the container, eight moles of hydrogen and two moles of nitrogen.

The next step is to determine how many moles of ammonia are created in the reaction, and if there is any excess reactant left over after the reaction. Since there is a 1:3 ratio for hydrogen gas to nitrogen gas, only six of the moles of hydrogen gas will be used in order to react will all two moles of the nitrogen gas.

$2mol\ N_2*\frac{3mol\ H_2}{1mol\ N_2}=6mol\ H_2$

This leaves two moles of excess hydrogen gas. Using stoichiometry and the molar ratios, we determine that four moles of ammonia are created in the reaction that comsumes two moles of nitrogen.

$2mol\ N_2*\frac{2mol\ NH_3}{1mol\ N_2}=4mol\ NH_3$

Four moles of ammonia, plus the two remaining moles of hydrogen gas, results in six moles of total gas after the reaction has run to completion.

Six moles is 60% of the inital moles in the container, so the final pressure will be 60% of the initial pressure. We can solve using the ideal gas law.

$\frac{P_1V_1}{n_1T_1}=\frac{P_1V_1}{n_1T_1}\rightarrow \frac{P_1}{n_1}=\frac{P_2}{n_2}$

$\frac{5atm}{10mol}=\frac{P_2}{6mol}$

A glass container holds a mixture of two gases. Gas A exerts a pressure of 5atm on the container.

If there are twice as many moles of gas A as there are moles of gas B in the container, what is the total pressure in the container?

The pressure is unaffected by the gas of lower concentration

Explanation

This is an equation which requires us to find the partial pressures of each gas in the container. The equation for partial pressure is $P_{a} = X_{a}P_{total}$ where is the molar fraction of one of the gases. Since there are twice as many moles of gas A as there are gas B, we know that gas A accounts for 2/3 of the moles in the container.

$X_a=\frac{mol\ A}{total\ mol}=\frac{2mol\ A}{(2mol A)+(1molB)}=\frac{2}{3}$

$5 = \frac{2}{3}P_{total}$

$P_{total} = 7.5atm$

How many moles of Oxygen gas are present at a volume of 10 L at 1 atm and 25o C? (MW Oxygen gas = 32 g/mol)

4 mol

0.041 mol

0.41 mol

14 mol

2.5 mol

Explanation

use PV = nRT

n = PV/ RT

P = 1 atm; V = 10 L; R = 0.0821 Latm/molK; T = 298 K (MUST switch temperature to K)

n = moles of gas

n = (1atm)(10L)/(0.0821Latm/molK)(298K)

n= 0.41 mol

An ideal gas takes up 60L at 2 atm. If the gas is compressed to 30L, what will the new pressure be?

Explanation

Ideal gas law (modified)

P1V1 = P2V2

P1 = 2 atm; V1 = 60L; P2 = ?; V2 = 30L

(2)(60) = (X)(30)

P2 = 4 atm

An ideal gas is initially at the following conditions:

The gas is then isothermally compressed to 15 atm.

What is the volume of the gas after compression? Round to the nearest liter.

Explanation

The system is isothermal, which means the temperature is constant.

$P_{i}V_{i}=nRT=P_{f}V_{f}$

$V_{f}=\frac{P_{i}V_{i}}{P_{f}}=\frac{10:atm*35:L}{15:atm}=23.33:L\rightarrow23:L$

Each of the following compounds are contained in a closed container with a volume of 100mL. Which of the following will exert the most pressure?

60g CO2

50g O2

46g N2

54g F2

40g H2

Explanation

The pressure exerted is dependent on molar concentration only (not mass). As the containers are all the same volume, the container with the most moles will exert the most pressure. 60g of 44g/mole CO2 gives 1.4mol, 50g of O2 gives 1.4mol, 46g N2 gives 1.6mol, and 54g F2 gives 1.4mol, 40g H2 gives 40mol.

What is the final pressure of a gas initially has a pressure of 10 atm at 50 L if the volume s now 25 L?

20 atm

5 atm

10 atm

25 atm

50 atm

Explanation

Use P1V1 = P2V2

P1 = 10atm; V1 = 50L

P2 = X; V2 = 25L

(10atm)(50L) = (x)(25L)

500 = 25x

x = 20

A sealed container containing 15L of oxygen gas, $O_{2}_{(g)}$ , with a pressure of 3atm at $\small 25^{o}C$ . How many grams of oxygen gas is present in the container? Assume ideal behavior of the gas.

$\small R= 0.08206\frac{L\cdot atm}{mol \cdot K}$

$\small 58.9 g$

$\small 1.85 g$

$\small 0.0185 g$

$\small 21.9 g$

Explanation

$\small PV = nRT$

$\small P$ is pressure, $\small V =$ volume, $\small n =$ moles, $\small R =$ gas constant, $\small T =$ temperature. Rearrange the equation, plug in the appropriate values and solve.