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MCAT Physical : Gas Laws

Study concepts, example questions & explanations for MCAT Physical

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Example Questions

Example Question #11 : Gas Laws

What is the temperature of a 1L container at STP after the pressure is doubled?

Possible Answers:

546K

273K

298K

200K

596K

Correct answer:

546K

Explanation:

Using Gay-Lussac's Law, which is $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ , we can find the change in temperature when the pressure is doubled.

Because the problem states that the original conditions were at STP, we know that pressure is 1atm and temperature is 273K. Since pressure and temperature are directly proportional, doubling pressure will also double the temperature. The final temperature will be 546K.

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Example Question #472 : Mcat Physical Sciences

$\small 1.8L$ of carbon dioxide at an original temperature of $\small 20^oC$ is heated to $\small 40^oC$ . What is the new volume of the gas?

Possible Answers:

1.68L

1.9L

3.6L

0.90L

Correct answer:

1.9L

Explanation:

Remember to convert temperature to Kelvin when using the gas equations.

20^oC+273=293K

40^oC+273=313K

We will use Charles's Law to calculate the new volume:

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

Use the given temperatures and initial volume to calculate the final volume.

$\frac{1.8 L}{293 K} = \frac{V_{2}}{313 K}$

$V_{2} = \frac{(1.8 L)(313 K)}{293 K} = 1.9 L$

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Example Question #12 : Gas Laws

Which of the following would not cause a decrease in the pressure of a gas in a sealed container?

Possible Answers:

Increasing the volume of the container

Removing gas particles from the container

Reducing the temperature

Adding moles of a different gas

Correct answer:

Adding moles of a different gas

Explanation:

A decrease in pressure means a decrease in gas particle collisions. The only option that would not cause a decrease in collisions is adding moles of a different gas. Even though different molecules are added, there will be greater pressure as particle collisions will be more frequent.

Reducing temperature slows the gas particles, thus decreasing the frequency of collisions. Similarly, increasing the volume of the container and removing particles will cause a decrease in collisions, and subsequent pressure.

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Example Question #13 : Gas Laws

A sealed container holds three moles of gas at 1atm and 200K. Its pressure is to 2atm. What will be the resulting temperature in the container?

Possible Answers:

273K

300K

250K

100K

400K

Correct answer:

400K

Explanation:

In the problem, the volume and the number of moles are constant and the temperature and pressure are the only two variables that are changing. Using the ideal gas law we can find that temperature and pressure are directly proportional. When pressure increases by a factor of two, temperature will also increase by a factor of two.

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

$\frac{(1atm)V}{200K}=\frac{(2atm)V}{T_2}$

T_2=400K

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Example Question #14 : Gas Laws

A $\small 10L$ container holds $\small 53mol$ of oxygen gas at a temperature of $\small 100^oC$ . The temperature remains constant and the volume of the container is increased to $\small 30L$ . What is the final pressure of the gas in terms of the initial pressure, $\small P_1$ ?

Possible Answers:

3P_1

$\frac{1}{20}P_1$

P_1

$\frac{1}{3}P_1$

Correct answer:

$\frac{1}{3}P_1$

Explanation:

The amount of gas is irrelevant. If the temperature is held constant and the volume is increased by a factor of three, the resulting pressure is decreased by a factor of three according to Boyle’s Law.

P_1V_1=P_2V_2

P_1(10L)=P_2(30L)

$P_2=\frac{P_1(10L)}{30L}$

$P_2=P_1(\frac{1}{3})$

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Example Question #51 : Fluids And Gases

A gas with an initial volume of $\small 0.2L$ and a temperature of $\small 290K$ is at a pressure of $\small 50kPa$ . What is the new pressure if the volume is increased to $\small 0.4L$ and the temperature is increased to $\small 400K$ ?

Possible Answers:

34.5kPa

42.8kPa

52.1kPa

58.4kPa

25.7kPa

Correct answer:

34.5kPa

Explanation:

To solve this problem we use the combined gas law:

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

Use the given values for the temperature and volume, as well as the initial pressure, to solve for the final pressure.

$\frac{(50 kPa)(0.2 L)}{290K} = \frac{P_{2}(0.4 L)}{400 K}$

$P_{2} = \frac{(50 kPa)(0.2 L)(400K)}{(290 K)(0.4 L)} = 34.5 kPa$

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MCAT Physical : Gas Laws

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #11 : Gas Laws

Example Question #472 : Mcat Physical Sciences

Example Question #12 : Gas Laws

Example Question #13 : Gas Laws

Example Question #14 : Gas Laws

Example Question #51 : Fluids And Gases

All MCAT Physical Resources

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