Question:
A sample of gas with $\gamma=1.5$ is taken through an adiabatic process in which the volume is compressed from $1200 \mathrm{~cm}^{3}$ to $300 \mathrm{~cm}^{3}$. If the initial pressure is $200 \mathrm{kPa}$. The absolute value of the workdone by the gas in the process $=$
Solution:
$v=1.5$
$\mathrm{p}_{1} \mathrm{v}_{1}^{\mathrm{v}}=\mathrm{p}_{2} \mathrm{v}_{2}^{\mathrm{v}}$
$(200)(1200)^{1.5}=\mathrm{P}^{2}(300)^{1.5}$
$\mathrm{P}_{2}=200[4]^{3 / 2}=1600 \mathrm{kPa}$
$\mid$ W.D. $\mid=\frac{\mathrm{p}_{2} \mathrm{v}_{2}-\mathrm{p}_{1} \mathrm{v}_{1}}{\mathrm{v}-1}=\left(\frac{480-240}{0.5}\right)=480 \mathrm{~J}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.