First law of thermodynamics: An ideal gas with γ = 1.30 occupies 7.0 L at 300 K and 200 kPa pressure. It is compressed adiabatically to 1/7 of its original volume, then cooled at constant volume to 300 K, and finally allowed to expand isothermally to 7.0 L. How much work does the gas do during this process? The ideal gas constant is R = 8.314 J/mol ∙ K = 0.0821 L ∙ atm/mol ∙ K.

Question

Quizplus · Accepted Answer

The work done by the gas during the adiabatic compression is given by:$$W_{ad} = -\int_{V_i}^{V_f} PdV$$where $P$ is the pressure, $V$ is the volume, and $i$ and $f$ denote the initial and final states. For an adiabatic process, $PV^\gamma = constant$, so we have:$$W_{ad} = -\int_{V_i}^{V_f} \frac{nRT}{V^\gamma}dV = -\frac{nRT}{\gamma-1}\left[\frac{1}{V_f^{\gamma-1}} - \frac{1}{V_i^{\gamma-1}}ight]$$where $n$ is the number of moles of gas, $R$ is the ideal gas constant, and $T$ is the temperature. Substituting the given values, we get:$$W_{ad} = -\frac{(1 	ext{ mol})(8.314 	ext{ J/mol} \cdot 	ext{K})(300 	ext{ K})}{1.30-1}\left[\frac{1}{(1/7 	ext{ L})^{1.30-1}} - \frac{1}{(7.0 	ext{ L})^{1.30-1}}ight] = -980 	ext{ J}$$The work done by the gas during the constant-volume cooling is zero, since the volume does not change. The work done by the gas during the isothermal expansion is given by:$$W_{iso} = -nRT\ln\left(\frac{V_f}{V_i}ight) = -(1 	ext{ mol})(8.314 	ext{ J/mol} \cdot 	ext{K})(300 	ext{ K})\ln\left(\frac{7.0 	ext{ L}}{1/7 	ext{ L}}ight) = 6400 	ext{ J}$$Therefore, the total work done by the gas during the process is:$$W_{total} = W_{ad} + W_{iso} = -980 	ext{ J} + 6400 	ext{ J} = 5420 	ext{ J}$$

First Law of Thermodynamics: an Ideal Gas with γ =