Ideal and Real Gases Questions and Answers

Answer: C

The critical compressibility factor is \(\frac{3}{8}\).

For the van der Waals equation, the conditions \((\partial P/\partial V)_{T} = 0\) and \((\partial^{2}P/\partial V^{2})_{T} = 0\) at the critical point give

\[ \begin{aligned} V_{c} &= 3b, \quad T_{c} = \frac{8a}{27Rb} \\ P_{c} &= \frac{a}{27b^{2}} \end{aligned} \]

\[ \begin{aligned} Z_{c} &= \frac{P_{c}V_{c}}{RT_{c}} \\ &= \frac{(a/27b^{2})(3b)}{8a/27b} = \frac{3}{8} \end{aligned} \]

The value 0.375 is the same for every van der Waals gas, higher than the 0.23 to 0.31 measured for real gases. The fractions 2/7, 5/8 and 1/7 do not come out of these critical constants.

Answer: C

The valid option is only Q and R are correct.

• P is false: in a reversible adiabatic process the temperature changes, and for an ideal gas dh = c_pdT, so the enthalpy changes.
• Q is true: ds = δq_rev/T, and with no heat transfer in a reversible process the entropy change is zero (the process is isentropic).
• R is true: the enthalpy of an ideal gas depends on temperature alone, so an isothermal process leaves it unchanged.
• S is false: a reversible isothermal process exchanges heat, so Δs = q/T is not zero (it is R ln(V₂/V₁) per mole).

The options naming P or S therefore cannot be right.

Answer: D

The molar volume of propane is 0.977 m³/kmol.

By the principle of corresponding states, gases at the same reduced pressure and temperature have the same compressibility factor, so propane also has Z = 0.9. Its actual conditions are

\[ \begin{aligned} T &= 1.5 \times 369.8 = 554.7\ \text{K} \\ P &= 1.0 \times 42.48\ \text{bar} = 4248\ \text{kPa} \end{aligned} \]

\[ \begin{aligned} \bar{v} &= \frac{ZRT}{P} = \frac{0.9 \times 8.314 \times 554.7}{4248} \\ &= 0.977\ \text{m}^{3}\text{/kmol} \end{aligned} \]

Treating propane as an ideal gas (Z = 1) would give 1.086, and 0.355, 0.526 and 0.791 do not follow from Z = 0.9 at these conditions.

Answer: A

The final pressure is 130.7 kPa.

The cooling happens in two stages. First the piston descends at constant pressure, 150 kPa, until it lands on the stops, which halves the volume (from 2 m to 1 m of height):

\[ T_{stop} = 673.15 \times \tfrac{1}{2} = 336.6\ \text{K} \]

That is still above 293.15 K, so the air keeps cooling at constant volume with the piston on the stops:

\[ p_{2} = 150 \times \frac{293.15}{336.6} = 130.7\ \text{kPa} \]

150 kPa would be right only if the piston never reached the stops. The higher pressures would require heating, not cooling.

Answer: B

The final temperature is -182°C.

For a reversible adiabatic process \(TV^{\gamma - 1}\) is constant:

\[ \begin{aligned} T_{2} &= T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma - 1} = 273\left(\frac{1}{81}\right)^{0.25} \\ &= \frac{273}{3} = 91\ \text{K} \end{aligned} \]

since \(81^{0.25} = 3\). In Celsius, 91 - 273 = -182°C.

-91°C comes from writing 91 K as -91°C without converting, and 0°C would mean no cooling, which an adiabatic expansion cannot give. -235°C does not follow from the adiabatic relation.

Answer: B

The average kinetic energy becomes 2E.

The average translational kinetic energy of a gas molecule is \(\tfrac{3}{2}kT\), proportional to the absolute temperature.

\[ \begin{aligned} \frac{E_{2}}{E_{1}} &= \frac{327 + 273}{27 + 273} \\ &= \frac{600}{300} = 2 \end{aligned} \]

The options 27E and 327E wrongly use Celsius temperatures, and the ratio 327/27 (about 12) is meaningless physically. E would be correct only if the temperature did not change.

Answer: A

The gas constant is 284.6 J/kg K.

For an ideal gas \(R = c_{p} - c_{v}\) and \(\gamma = c_{p}/c_{v}\), so

\[ \begin{aligned} R &= c_{p}\left(1 - \frac{1}{\gamma}\right) = 1147 \times \frac{0.33}{1.33} \\ &= 284.6\ \text{J/kg K} \end{aligned} \]

862.4 J/kg K is \(c_{v} = 1147/1.33\), not \(R\). 1005 J/kg K is the usual \(c_{p}\) of air, a different gas from this one, and 8314 J/kmol K is the universal gas constant, which still has to be divided by the molar mass.

Answer: D

The specific work done on the gas is 63.3 kJ/kg.

For a reversible adiabatic process \(pv^{\gamma}\) is constant, so the final specific volume is

\[ \begin{aligned} v_{2} &= v_{1}(p_{1}/p_{2})^{1/\gamma} \\ &= 0.45 \times (98/480)^{0.714} \\ &= 0.1446\ \text{m}^{3}/\text{kg} \end{aligned} \]

The work done on the gas in a closed (non-flow) adiabatic compression is

\[ \begin{aligned} w &= \frac{p_{2}v_{2} - p_{1}v_{1}}{\gamma - 1} \\ &= \frac{480(0.1446) - 98(0.45)}{0.4} \\ &= \frac{69.4 - 44.1}{0.4} = 63.3\ \text{kJ/kg} \end{aligned} \]

25.4 is close to the numerator \(p_{2}v_{2} - p_{1}v_{1}\) (25.3) left undivided by \(\gamma - 1\), and 12.6 and 18.5 are smaller still; none of them accounts for the full adiabatic work.

Answer: B

The entropy change is \(mR\ln(P_{1}/P_{2})\).

For an ideal gas:

\[ \Delta S = mc_p\ln\frac{T_2}{T_1} - mR\ln\frac{P_2}{P_1} \]

The process is isothermal, so the temperature term vanishes and \(\Delta S = mR\ln(P_{1}/P_{2})\). Since heat is lost, the gas is being compressed (P₂ > P₁) and \(\Delta S\) is negative, as it must be for a reversible process with heat leaving.

\(mR\ln(P_{2}/P_{1})\) has the wrong sign. Subtracting Q/T₁ counts the heat transfer twice, because for a reversible process the expression already equals the heat transfer over T. Zero applies to a reversible adiabatic process, not to an isothermal one.

Answer: B

The FALSE statement is the one claiming that a real gas in an adiabatic reversible process follows \(PV^{\gamma}\) = constant.

The relation \(PV^{\gamma}\) = constant is derived from the first law using \(PV = mRT\) and constant specific heats, so it holds for an ideal (perfect) gas only. A real gas, whose equation of state and specific heats differ from the ideal ones, does not follow it in general.

The other statements are true. An ideal gas's enthalpy depends on temperature alone, so it is independent of pressure. The equation \(P/R = mT/V\) is just \(PV = mRT\) rearranged; the ideal gas equation holds at every state, whatever polytropic process (here \(PV^{1.5}\) = constant) connects them. And any real gas approaches ideal behaviour at low pressure or high temperature, where molecular volume and attraction become negligible.