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GRE Subject Test: Chemistry : Gas Laws

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Example Questions

Example Question #22 : Thermodynamics And Phases

A gas sample is contained in a 4L vessel at a pressure of 3atm. Assuming all other conditions are kept constant, what is the new pressure in the vessel if the volume is reduced to 1.5L?

Possible Answers:

4.5 atm $\displaystyle 4.5 atm$

8.0 atm $\displaystyle 8.0 atm$

$\displaystyle 1.12 atm$

6.0 atm $\displaystyle 6.0 atm$

Correct answer:

8.0 atm $\displaystyle 8.0 atm$

Explanation:

According to Boyle's law, pressure and volume are inversely proprotional to each other. This is represented by the equation:

$P_{1}V_{1} = P_{2}V_{2}$ $\displaystyle P_{1}V_{1} = P_{2}V_{2}$

In other words, as volume decreases in a vessel, the pressure will increase, and vice versa. Using the given conditions, we can solve for the final pressure in the vessel:

$(4L)(3atm) = (1.5L)P_{2}$ $\displaystyle (4L)(3atm) = (1.5L)P_{2}$

$P_{2} = 8.0 atm$ $\displaystyle P_{2} = 8.0 atm$

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Example Question #23 : Thermodynamics And Phases

An unknown amount of neon gas is contained in a 3.00L vessel. At a temperature of $40.0^{\circ} C$ $\displaystyle 40.0^{\circ} C$ , the gas exerts a pressure of 4.00atm.

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

Neon gas has a molar mass of $20.18\frac{g}{mol}$ $\displaystyle 20.18\frac{g}{mol}$ .

Based on these conditions, what is the mass of neon gas in the vessel?

Possible Answers:

0.46 g $\displaystyle 0.46 g$

73.7 g $\displaystyle 73.7 g$

16.2 g $\displaystyle 16.2 g$

9.42 g $\displaystyle 9.42 g$

Correct answer:

9.42 g $\displaystyle 9.42 g$

Explanation:

This question deals with the amount of gas present in a vessel for only one set of conditions. This makes the ideal gas law a suitable equation to use in order to determine the amount of gas in the vessel. The ideal gas law is written as:

$\displaystyle PV = nRT$

Using this equation, we can solve for the molar quantity of gas in the vessel:

$\displaystyle PV = nRT$

$(4atm)(3L) = n(0.0821\frac{L\cdot atm}{mol\cdot K})(313K)$ $\displaystyle (4atm)(3L) = n(0.0821\frac{L\cdot atm}{mol\cdot K})(313K)$

$\displaystyle n = 0.467 mol$

Knowing this, we can now solve for the mass of the gas in the vessel by multiplying this molar amount by the molar mass:

$Moles = \frac{mass}{molar\ mass}$ $\displaystyle Moles = \frac{mass}{molar\ mass}$

$0.467 = \frac{mass}{20.18\frac{g}{mol}}$ $\displaystyle 0.467 = \frac{mass}{20.18\frac{g}{mol}}$

$\displaystyle mass = 9.42 g$

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Example Question #22 : Thermodynamics And Phases

Which gas follows the exact definition of the ideal gas law?

Possible Answers:

$O_{2}$ $\displaystyle O_{2}$

$\displaystyle He$

$NH_{3}$ $\displaystyle NH_{3}$

None of these

$N_{2}$ $\displaystyle N_{2}$

Correct answer:

None of these

Explanation:

Though the ideal gas law gives a nearly close to real approximation of numbers, it oversimplifies its description of gases. No real gas follows the exact definition of the ideal gas law and is very complex because there are intermolecular forces that must be considered. An ideal gas described as a point mass in which the particles are so small that its volume is negligible. However, real gases have real volume. Also, ideal gases are considered elastic, having no attractive and repulsive forces with no energy transfer during collisions. Real gases actually collide and are non-elastic. Note that gases approach ideal behavior as their temperature increases and their pressure decreases.

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GRE Subject Test: Chemistry : Gas Laws

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #22 : Thermodynamics And Phases

Example Question #23 : Thermodynamics And Phases

Example Question #22 : Thermodynamics And Phases

All GRE Subject Test: Chemistry Resources

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