Solve this following

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. $\mid=\frac{\mathrm{p}_{2} \mathrm{v}_{2}-\mathrm{p}_{1} \mathrm{v}_{1}}{\mathrm{v}-1}=\left(\frac{480-240}{0.5}\right)=480 \mathrm{~J}$

 

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