Question

A monoatomic gas is taken through cyclic process ABCA where AB is an isothermal process. Volume of gas is in liters and pressure is in kilo Pascals. Temperature at A is 300 K, then [Use ln(2)=0.693]

• A. Temperature at C is 75 K.
• B. Heat absorbed in process AB is 16ln(2) J.
• C. Work done in full cycle ABCA is 17 J.
• D. Heat absorbed in full cycle ABCA is 9 J.

Answer 1

Answer: (A), (B)

The correct options are
A Temperature at C is 75 K.
B Heat absorbed in process AB is 16ln(2) J.

By $\frac{P_A}{T_A}=\frac{P_C}{T_C}\Rightarrow T_C=\frac{300}{4}=75K$Work done in the cycle is $W=W_{AB}+W_{BC}+W_{CA}{W}_{BC}=(1-4)\times {10}^{-3}\times 2\times {10}^3=-6J{W}_{AB}=nRT\ln \frac{V_2}{V_1}=8\times {10}^3\times {10}^{-3}\ln \left(4\right)W_{CA}=0\Rightarrow W=16\ln \left(2\right)-6=5J$By
$\Delta Q=\Delta U+\Delta W\Rightarrow \Delta Q_{AB}=\Delta U_{AB}+\Delta W_{AB}=0+16\ln \left(2\right)(Since\Delta T=0,\Delta U_{AB}=0)$