Question

A gas is taken through a cyclic process ABCA as shown in figure. If 2.4 cal of heat is given in the process, what is the value of J ?

Answer 1

$\Delta \text{Q}=2.4\text{cal}$

$\Delta \text{W}=W_{\text{AB}}+W_{\text{BC}}+W_{\text{AC}}$

$=0+\frac{1}{2}\times (100+200)\times {10}^3\times 200\times {10}^{-6}-100\times {10}^3\times 200\times {10}^{-6}$

$=\frac{1}{2}\times 300\times {10}^3\times 200\times {10}^{-6}-20$

$=30\text{J}-20\text{J}=10\text{J}$

$\Delta \text{U}=0$ (in a cyclic process)

$\Delta \text{Q}=\Delta \text{U}+\Delta \text{W}$

$\Rightarrow 2.4\text{J}=10$

$\Rightarrow J=\frac{10}{2.4}=\frac{100}{24}=\frac{25}{6}=4.17\text{J}/\text{cal}$