One mole of an ideal monoatomic gas undergoes | vedclass

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(D) For $A \rightarrow B$ (Reversible adiabatic process): $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$ Given: $T_1 = 600 \ K$,$T_2 = 60 \ K$,$V_1 = 10 \

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) For $A ightarrow B$ (Reversible adiabatic process): $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$ Given: $T_1 = 600 \ K$,$T_2 = 60 \ K$,$V_1 = 10 \ m^3$,and $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = 5/3$. $600 	imes (10)^{5/3 - 1} = 60 	imes (V_2)^{5/3 - 1}$ $10 = (V_2 / 10)^{2/3}$ $\Rightarrow 10^{3/2} = V_2 / 10$ $\Rightarrow V_2 = 10^{5/2}$. For the total process: $q_{total} = q_{AB} + q_{BC} = R T_2 \ln 10$. Since $A ightarrow B$ is adiabatic,$q_{AB} = 0$. $q_{BC} = n R T_2 \ln(V_3 / V_2) = 1 	imes R 	imes 60 	imes \ln(V_3 / 10^{5/2}) = 60 R \ln 10$. $\ln(V_3 / 10^{5/2}) = \ln 10 \Rightarrow V_3 / 10^{5/2} = 10$. $V_3 = 10 	imes 10^{5/2} = 10^{7/2}$. Taking $\log$ on both sides: $\log V_3 = 7/2 \log 10 = 3.5$. Therefore,$2 \log V_3 = 2 	imes 3.5 = 7$.

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