Question

The temperature of a one mole of diatomic gas changes from 4T to T in adiabatic process. If R is universal gas constant. Then work done

• A. $\frac{15RT}{3}$
• B. $\frac{15RT}{2}$
• C. $\frac{3RT}{15}$
• D. $\frac{12RT}{5}$

Answer 1

The correct option is B $\frac{15RT}{2}$

For a diatomic gas, the value of $\gamma$ is 1.4
For this process, the work done in the cycle will be,
$W=\frac{nR\Delta T}{\gamma -1}$
$\delta T=4T-T=3T$
Substitutng these values,
$W=\frac{15RT}{2}$