Fluids and Gases - MCAT Physical

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Question

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

Answer

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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Question

Which of the following factors does not explain why measurements of real gases deviate from ideal values?

Answer

Measurements of real gases deviate from ideal gas predictions because intermolecular forces and the volume of the particles themselves are not taken into consideration for ideal gases. The volume of the space between particles is considered for ideal gases and does not contribute to deviation from ideal gas behavior.

Attraction between molecules causes real pressure to be slightly less than ideal pressure, while the volume of gas particles causes real volume to be slightly greater than ideal volume.

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Question

8 liters of an ideal gas is in an isolated container at 30 degrees Celsius. The container is heated at constant pressure until its volume is doubled. What is the new temperature of the gas?

Answer

At constant pressure, $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$ , where the temperatures are measured in Kelvin (absolute temperature).

First, convert the given temperature (C) to Kelvin (K).

K = C + 273 = 30 + 273 = 303K

Plug the temperature and volumes into the above equation and solve for the final temperature.

$\frac{8 liters}{303 K}=\frac{16 liters}{T_{2}}$

$T_{2} = 606 K$

Convert this value back to Celsius.

C = K - 273 = 606 - 273 = 333oC

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Question

Five moles of nitrogen gas are present in a 10L container at 30oC. What is the pressure of the container?

Answer

Using the ideal gas law equation we can find that P= nRT/V. We then plug in the given values.

$P=\frac{5mol0.0821\frac{Latm}{molK}303K}{10L}$

Solving for P gives us 12.4atm.

Note: 30oC must be converted into Kelvin by adding 273K

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Question

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

Answer

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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Question

Regarding the following sets of conditions, which answer option gives the correct listing of systemic pressures from greatest to least?

$\text{I.}\ 2mol N_2,\ 3mol O_2;\ T=273K;\ V=10L$

$\text{II.}\ 5mol N_2;\ T=546K;\ V=5L$

$\text{III.}\ 2.5mol N_2;\ T=273K;\ V=10L$

$\text{IV.}\ 5mol N_2,\ 5mol O_2;\ T=546K;\ V=5L$

Answer

This question asks for you to look at a set of conditions for gases, and determine relative pressures. The best equation to use for quick calculation and relation is the ideal gas law, given by:

Rearranging this, and removing the constant (since it will not affect relative pressure), we can generate a proportionality of pressure to the other variables.

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

We can use this proportionality with each option to determine their rankings by pressure.

$P_I=\frac{(5mol)(273K)}{10L}$

$P_{II}=\frac{(5mol)(546K)}{(5L)}=\frac{2(5mol)(273K)}{\frac{1}{2}(10L)}=4P_I$

$P_{III}=\frac{(2.5mol)(273K)}{10L}=\frac{\frac{1}{2}(5mol)(273K)}{10L}=\frac{1}2{}P_I$

$P_{IV}=\frac{(10mol)(546K)}{5L}=\frac{2(5mol)2(273K)}{\frac{1}{2}(10L)}=8P_1$

$\frac{1}{2}P_I<P_I<4P_I<8P_I$

$P_{III}<P_I<P_{II}<P_{IV}$

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Question

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?

Answer

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.

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

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

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Question

If one mole of oxygen gas occupies a 5L container at 300K, what is the pressure in the container?

$R=0.0821\frac{atmL}{molK}$

Answer

Using the ideal gas law, , we can solve for pressure.

$P(5L)= (1mol)(0.0821 \frac{atmL}{mol K})(300K)$

$P= \frac{(1mol)(0.0821\frac{atmL}{molK})(300K)}{5L}=\frac{24.63atm*L}{5L}=4.9atm$

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Question

Recall that the ideal gas law states that , where $\frac{L \cdot atm}{mol \cdot K}$ .

If there are 5g of hydrogen gas in a 10L chamber at 32°C, what would be the pressure?

Answer

Using the equation and solving for P you get, $P = \frac{nRT}{V}$ .

Recall that hydrogen forms a diatomic molecule when in gas form. This should always be an assumption when working with hydrogen gas on the MCAT. When we convert 5g to moles, we must use a conversion factor of 2g/mol.

$5g\ H_2\frac{1mol\ H_2}{2g\ H_2}=2.5mol\ H_2$

Temperature must be converted to Kelvin. You must have this conversion memorized for the MCAT.

Now we can solve for P.

$P=\frac{nRT}{V}=\frac{(2.5mol)(0.0821\frac{Latm}{mol*K})(305K)}{10L}=6.26atm$

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Question

A certain gas is initially at a pressure of 2atm in a volume of 5L. It then experiences a decrease in volume to 2L, and is held at a constant temperature throughout the process. What is the new pressure?

Answer

Since this is an isothermal change (constant temperature), this falls under Boyle's law.

Taking this equation and solving for the new pressure (P2) we come up with 5atm.

$P_2=\frac{P_1V_1}{V_2}=\frac{(2atm)(5L)}{(2L)}=5atm$

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Question

15L of a gas is held at constant pressure as the temperature increases from 300K to 350K. What is the new volume?

Answer

Charles's Law states that $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ . To solve for the final volume, we simply plug in our given values to this equation.

$V_2=\frac{V_1T_2}{T_1}=\frac{(15L)(350K)}{(300K)}=17.5L$

This law applies only for isobaric (constant pressure) changes.

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Question

A balloon filled with one mole of an ideal gas is leaking molecules at a constant rate of $\small0.01\frac{mol}{hr}$ .

After 75 hours, the pressure is half of the initial pressure. What is the new volume in terms of the initial volume, $\small V_i$ ?

Assume temperature remains constant.

Answer

We can use the given values to determine how many moles of the gas have leaked out after 75 hours.

$0.01\frac{mol}{hr}*75hr=0.75mol$

We now know that the balloon started with one mole, and that 0.75 moles have leaked. This means that 0.25 moles remain in the balloon. We now have our initial and final mole values and pressure values. We can rearrange the ideal gas law to isolate the variables we need, assuming that the temperature is constant.

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

$\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}$

Using our proportions, we can try to solve for the final volume. For simplicity, assume the initial pressure is 1 and the final pressure is 0.5.

$\frac{(1)V_1}{1}=\frac{(0.5)V_2}{0.25}$

$V_1=2V_2\rightarrow V_2=\frac{1}{2}V_1$

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Question

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, $\small P_1$ ?

Answer

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.

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

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

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Question

$\small 1.8L$ 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?

Answer

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

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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Question

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 $\small 400K$ ?

Answer

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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Question

How many moles of oxygen gas are in a $\small 560mL$ sample that is at a temperature of $\small 75^oC$ and a pressure of $\small 2.1atm$ ?

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

Answer

First, each value must be converted to the correct units given for the gas constant.

$560 mL *\frac{1 L}{1000 mL} = 0.56 L$

Next, use the ideal gas law to solve for moles.

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

$n = \frac{(2.1 atm)(0.56 L)}{(0.0821)(348 K)} = 0.041 mol$

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Question

How many grams of ammonia gas are in a $\small 8.0L$ sample at $\small 30.0^oC$ and $\small 3.0atm$ ?

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

Answer

First, convert temperature to Kelvin.

Next, use the ideal gas law to solve for moles.

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

$n = \frac{(3 atm)(8.0 L)}{(0.0821)(303K)} = 0.96 mol$

Finally, convert moles of ammonia to grams using molar mass.

$0.96mol\ NH_3*\frac{17.0g\ NH_3}{1mol\ NH_3}=16g\ NH_3$

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Question

Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.

In section 3, as the temperature increases, the velocity of the molecules in a(n) manner.

Answer

This question asks you how the velocity of the molecules changes with increasing temperature. From molecular kinteics, we know that temperature is proportional to kinetic energy.

$Kinetic\ Energy=\frac{3}{2}KT$

We can see that as temperature rises, the kinetic energy rises, however, the question asks us how the velocity changes. Now, we need to know how kinetic energy and velocity are related. We can pull this relationship from Newtonian mechanics.

$Kinetic\ Energy=\frac{3}{2}KT=\frac{1}{2}mv^2$

If we simplify, we can see that $V=\sqrt{\frac{3KT}{m}}$ . Velocity increases with increasing temperature, but in an indirectly proportional manner.

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Question

At room temperature which of the following equimolar gases has the fastest effusion rate?

Answer

According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of the mass: $Rate \propto \frac{1}{\sqrt{mass}}$ .

The gas with the lowest molecular weight will effuse the fastest. has the lowest molecular weight of the given gasses, with a value of $17 \frac{g}{mol}$ .

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Question

Consider a helium atom and a molecule of carbon monoxide both at a temperature of $\small 480K$ . Which of the following accurately describes the velocity of the carbon monoxide molecule in relation to the velocity of the helium atom?

Answer

Since both molecules are at the same temperature, the value of the temperature is irrelevant. The difference in velocity between the two molecules depends on the difference in their molar masses. The molar mass of a helium atom is $\small 4amu$ , and the molar mass of a carbon monoxide molecule is .

Graham's Law describes the relationship between gas particle mass and velocity:

$\frac{v_{1}}{v_{2}} = \sqrt{\frac{m_{2}}{m_{1}}}$

Using the molar masses of the particles in question, we can determine their relationship.

$\frac{v_{CO}}{v_{He}} = \sqrt{\frac{4 atm}{28 atm}} = \sqrt{\frac{1}{7}}$

$v_{CO}=v_{He}\sqrt{\frac{1}{7}}$

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