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State the four assumptions of the classical ideal gas model.
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The atoms have random instantaneous velocities. The volumes of individual atoms are negligible compared to the container volume. Collisions between atoms are elastic. The only appreciable forces on atoms occur during collisions, so no long-range forces act between them.
Explain why the internal energy of an ideal gas depends only on its temperature, not on its volume or pressure?
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The ideal gas model assumes no forces act between atoms except during brief collisions, so there is no inter-particle potential energy. All internal energy is the kinetic energy of the atoms. Because kinetic energy depends only on temperature, internal energy depends only on temperature.
Indicate whether a real gas at very high pressure is more likely or less likely to behave as an ideal gas than the same gas at low pressure? Justify your response.
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A real gas at very high pressure is less likely to behave ideally. At high pressure the atoms are closer together, so the volume of the atoms is no longer negligible compared to the container volume. Inter-molecular forces between nearby atoms also become significant, violating two assumptions of the ideal gas model.
Calculate the number of oxygen molecules in a room of volume 45 m³ at a pressure of 1.01 × 10⁵ Pa and a temperature of 20 °C.
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Equation used: $$PV=N{k}_{B}T ⇒ N=\frac{PV}{{k}_{B}T}$$ Converting temperature to kelvin: $$T=20+273=293 \text{K}$$ Given: $$P=1.01\times {10}^{5} \text{Pa}$$ $$V=45 {\text{m}}^{3}$$ $${k}_{B}=1.38\times {10}^{-23} \text{J/K}$$ Substituting: $$N=\frac{1.01\times {10}^{5}\times 45}{1.38\times {10}^{-23}\times 293}$$ $$N=\frac{4.545\times {10}^{6}}{4.0434\times {10}^{-21}}$$ $$N=1.12\times {10}^{27} \text{molecules}$$ This is the total number of gas molecules. Oxygen makes up about 21% of air, so the number of oxygen molecules is approximately $0.21\times 1.12\times {10}^{27}\approx 2.36\times {10}^{26}$.
A sealed container holds gas at 2.0 × 10⁵ Pa and 400 K. The gas is cooled to 200 K at constant volume. Determine the new pressure.
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At constant volume and constant amount of gas: $$\frac{{P}_{1}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}} ⇒ {P}_{2}={P}_{1}\times \frac{{T}_{2}}{{T}_{1}}$$ Given: $${P}_{1}=2.0\times {10}^{5} \text{Pa}, {T}_{1}=400 \text{K}, {T}_{2}=200 \text{K}$$ Substituting: $${P}_{2}=2.0\times {10}^{5}\times \frac{200}{400}$$ $${P}_{2}=1.0\times {10}^{5} \text{Pa}$$ Halving the absolute temperature at constant volume halves the pressure, because pressure is directly proportional to temperature.
Indicate whether doubling the number of moles of gas in a rigid sealed container at constant temperature will more than double, exactly double, or less than double the pressure? Justify your response.
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The pressure exactly doubles. From $P=\frac{nRT}{V}$, pressure is directly proportional to $n$ when $T$ and $V$ are constant. Doubling $n$ therefore doubles $P$.
Describe how a graph of pressure versus temperature in degrees Celsius can be used to determine absolute zero?
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Plot measured pressure values against temperature in °C at constant volume. The data form a straight line. Extrapolate the line to the temperature axis where pressure equals zero. The temperature at that intercept is absolute zero, approximately $-273 ^{\circ}\text{C}$.
A gas at constant volume has its kelvin temperature tripled. Predict what happens to the pressure and justify your answer using the ideal gas law.
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The pressure triples. From $P=\frac{nRT}{V}$, pressure is directly proportional to absolute temperature at constant $n$ and $V$. Tripling $T$ therefore triples $P$.
Sketch a graph of pressure versus volume for a fixed amount of ideal gas at constant temperature. Indicate whether the graph is linear and justify your answer.
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The graph is a smooth curve (hyperbola) in the first quadrant that decreases as volume increases. It is not linear because $P=\frac{nRT}{V}$, so $P$ is inversely proportional to $V$. A linear graph would require $P∝V$, which contradicts the ideal gas equation at constant $T$.
Indicate whether a $P$ vs $T$ graph for a gas in a container of volume ${V}_{1}$ will have a steeper or shallower slope than the same amount of gas in a container of volume ${V}_{2}$, where ${V}_{2}>{V}_{1}$? Justify your response.
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The graph for ${V}_{1}$ has a steeper slope. The slope of the $P$ vs $T$ graph at constant volume is $\frac{nR}{V}$. Because ${V}_{1}<{V}_{2}$, the fraction $\frac{nR}{{V}_{1}}$ is larger than $\frac{nR}{{V}_{2}}$, producing a steeper line.