IChO 1994 · Theory problem 7

Ideal-gas expansion and a refrigeration cycle

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7.1 When an ideal, monatomic gas expands reversibly from a volume \(V_1\) to a volume \(V_2\), a work

\[ w=-\int_{V_1}^{V_2}p\,dV \]

is performed on the system by the surroundings. In this equation, \(w\) is the work and \(p\) is the pressure of the gas.

Determine the performed work when one mole ideal gas expands isothermally from \(V_1=1.00\ \mathrm{dm^3}\) to \(V_2=20.0\ \mathrm{dm^3}\) at the temperature \(T=300.0\ \mathrm{K}\).

Given: The gas constant \(R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\).

7.2 Determine how much heat must be added to the gas during the process given under 7.1.

7.3 The gas will perform less work in an adiabatic expansion than in an isothermal expansion. Is this because the adiabatic expansion is characterized by (check the square you think is most important).

1 The volume of the gas is constant
2 The expansion is always irreversible
3 No heat is supplied to the gas

7.4 The cyclic process shown schematically in Figure 1 shows the four steps in a refrigeration system with an ideal gas as working medium. Identify the isothermal and adiabatic steps in the process. Here, \(T_{\mathrm H}\) and \(T_{\mathrm C}\) represent high and low temperature, respectively. Specify for each step whether it is adiabatic or isothermal.

Figure 1

Official English text of the problem.

Answers

  • 7.1: \(w=-7.47\ \mathrm{kJ}\), using the stated convention for work done on the gas.
  • 7.2: \(q=+7.47\ \mathrm{kJ}\); this heat must be added to the gas.
  • 7.3: Check square 3: “No heat is supplied to the gas.”
  • 7.4: \(1\rightarrow2\): isothermal expansion at \(T_{\mathrm C}\); \(2\rightarrow3\): adiabatic compression; \(3\rightarrow4\): isothermal compression at \(T_{\mathrm H}\); \(4\rightarrow1\): adiabatic expansion.

Solution

7.1

For an ideal gas,

\[ pV=nRT. \]

Since the expansion is isothermal, \(T\) is constant, so

\[ w=-\int_{V_1}^{V_2}\frac{nRT}{V}\,dV =-nRT\ln\!\left(\frac{V_2}{V_1}\right). \]

Substituting the given quantities,

\[ \begin{aligned} w &=-(1\ \mathrm{mol}) (8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}) (300.0\ \mathrm{K}) \ln\!\left(\frac{20.0\ \mathrm{dm^3}}{1.00\ \mathrm{dm^3}}\right)\\ &=-7471.96\ \mathrm{J}\\ &=\boxed{-7.47\ \mathrm{kJ}}. \end{aligned} \]

The volume ratio is dimensionless. The negative sign means that the gas does work on its surroundings: the work delivered by the gas is \(7.47\ \mathrm{kJ}\).

7.2

With work defined as work done on the gas, the first law is

\[ \Delta U=q+w. \]

For a monatomic ideal gas,

\[ \Delta U=\frac32 nR\Delta T. \]

The temperature remains constant, so \(\Delta U=0\). Therefore,

\[ q=-w=\boxed{+7.47\ \mathrm{kJ}}. \]

Thus \(7.47\ \mathrm{kJ}\) of heat must enter the gas to replace the energy transferred to the surroundings as work.

7.3

The correct choice is 3: No heat is supplied to the gas.

In an adiabatic process, \(q=0\). During expansion the gas does work, so its internal energy decreases:

\[ \Delta U=w<0. \]

Consequently, its temperature falls. Comparing reversible expansions from the same initial state to the same final volume, the adiabatic gas therefore has a lower pressure at each intermediate volume than the isothermal gas.

For a monatomic ideal gas, \(\gamma=C_p/C_V=5/3\). Hence, for \(V>V_1\),

\[ p_{\mathrm{ad}} =p_1\left(\frac{V_1}{V}\right)^{5/3} < p_1\left(\frac{V_1}{V}\right) =p_{\mathrm{iso}}. \]

The area under the pressure–volume curve, \(\int p\,dV\), is consequently smaller for the adiabatic expansion, so the gas delivers less work.

Option 1 is incorrect because the volume increases during expansion. Option 2 is incorrect because an adiabatic expansion can be reversible.

7.4

The lower curve through states 1 and 2 is the isotherm at \(T_{\mathrm C}\); the upper curve through states 3 and 4 is the isotherm at \(T_{\mathrm H}\). The connecting branches are adiabatic.

Step Identification Reasoning
\(1\rightarrow2\) Isothermal expansion at \(T_{\mathrm C}\) Volume increases along the cold isotherm. The gas absorbs heat while doing work, maintaining constant temperature.
\(2\rightarrow3\) Adiabatic compression Volume decreases without heat exchange. Work done on the gas raises its temperature from \(T_{\mathrm C}\) to \(T_{\mathrm H}\).
\(3\rightarrow4\) Isothermal compression at \(T_{\mathrm H}\) Volume decreases along the hot isotherm. The gas releases heat to maintain constant temperature.
\(4\rightarrow1\) Adiabatic expansion Volume increases without heat exchange. The gas does work and cools from \(T_{\mathrm H}\) to \(T_{\mathrm C}\).

These identifications also agree with the slopes: isotherms obey \(pV=\text{constant}\), whereas reversible adiabats obey \(pV^{5/3}=\text{constant}\) and are steeper at a given state.

The solution was checked against the official solution.