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Heat of Reaction

A monatomic i...

Question

A monatomic ideal gas of two moles is taken through a cyclic process starting for $A$ as shown in the figure. The volume ratios are $(V_{B} / V_{A}) = 2$ and $(V_{D} / V_{A}) = 4$ . If the temperature $T_{A}$ at $A$ is $27^{\circ} C$ , calculate:
(a) The temperature of the gas at point $B$ ,
(b) Heat absorbed or released by the gas in each process.
(c) The total work done by the gas during the complete cycle.
Express your answer in terms of the gas constant $R$ .

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Solution

Taking $V_{A} = V_{0}; V_{B} = 2 V_{0}; V_{D} = 4 V_{0}$
Process $A \to B$ : Isobaric process as $V / T = c o n s t a n t$
$V_{0} / (273 + 27) = (2 V_{0}) / T_{B} \Rightarrow T_{B} = 600 K = 327^{\circ} C$ .
$Q_{1} = n C_{P} △ T = 2 (5 / 2) R (600 - 300) = 1500 R (a b s o r b e d)$
Process $B \to C$ (Isothermal)
$Q_{2} = W_{2} = n R T_{B} l n (V_{2} / V_{1}) = 831.78 R (a b s o r b e d)$
Process $C \to D$ (Isochoric)
$Q_{3} = n C_{V} △ T = - 900 R (r e l e a s e d)$
Process $D \to A$ (Isothermal)
$Q_{4} = W_{4} = n R T_{A} l n (V_{2} / V_{1}) = - 831.78 R$
Total word done: $W = Q_{1} + Q_{2} + Q_{3} + Q_{4} = 600 R$ .

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