Question

# A 2.00 mol diatomic gas initially at $$300 \mathrm{K}$$ undergoes this cycle: It is (1) heated at constant volume to $$800 \mathrm{K},(2)$$ then allowed to expand isothermally to its initial pressure, (3) then compressed at constant pressure to its initial state. Assuming the gas molecules neither rotate nor oscillate, find (a) the net energy transferred as heat to the gas, (b) the net work done by the gas, and (c) the efficiency of the cycle.

Solution

## (a) Processes 1 and 2 both require the input of heat, which is denoted $$Q_{\mathrm{H}} .$$ Noting that rotational degrees of freedom are not involved, then, $$C_{V}=3 R / 2, C_{p}=5 R / 2,$$ and $$\gamma=5 / 3 .\\$$ We further note that since the working substance is an ideal gas, process 2 (being isothermal) implies $$Q_{2}=W_{2}$$. Finally, we note that the volume ratio in process 2 is simply $$8 / 3 .\\$$ Therefore,$$Q_{\mathrm{H}}=Q_{1}+Q_{2}=n C_{V}\left(T^{\prime}-T\right)+n R T^{\prime} \ln \dfrac{8}{3}\\$$which yields (for $$T=300 \mathrm{K}$$ and $$T^{\prime}=800 \mathrm{K}$$ ) the result $$Q_{\mathrm{H}}=25.5 \times 10^{3} \mathrm{J}\\$$(b) The net work is the net heat $$\left(Q_{1}+Q_{2}+Q_{3}\right) .$$ We find $$Q_{3}$$ from$$\text { Thus, } W=4.73 \times 10^{3} \mathrm{J}\\$$(c) we find that the efficiency is$$\varepsilon=\dfrac{|W|}{\left|Q_{\mathrm{H}}\right|}=\dfrac{4.73 \times 10^{3}}{25.5 \times 10^{3}}=0.185 \text { or } 18.5 \%\\$$Physics

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