Question

A gas $(y=1.5)$ is compressed in an adiabatic process, then its volume changes from $1600c{m}^2$ to $400c{m}^3$. If the initial pressure is $150kPa$, then calculate the final pressure, and the work done on the gas.

Answer 1

Given, $y=1.5;$
$V_1=1600c{m}^3=1600\times {10}^{-6}{m}^3=1.6\times {10}^{-3}{m}^3;$
$V_2=400c{m}^3=400\times {10}^{-6}{m}^3=0.4\times {10}^{-3}{m}^3;$
$P_1=150kPa$
or $P_1=1850\times {10}^{3}Pa=1.5\times {10}^{5}Pa;P_2=?$
$\because$ In an adiabatic process
$P_2{V}_2^{\gamma }=P_1{V}_1^{\gamma }$
$\therefore P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }$
$=1.5\times {10}^5\times {\left(\frac{1.6\times {10}^{-3}}{0.4\times {10}^{-3}}\right)}^{1.5}$
$=1.5\times {10}^5\times (4{)}^{1.5}$
$=1.5\times {10}^5\times 4\sqrt{4}$
$=1.5\times {10}^5\times 4\times 2$
$=1.5\times 8\times {10}^5=12.0\times {10}^5$
$P_2=1.2\times {10}^{6}Pa$
$=1200\times {10}^{3}Pa$
or $P_2=1200kPa$
Work done in adiabatic process
$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$
$=\frac{150\times {10}^3\times 1.6\times {10}^{-3}-1200\times {10}^3\times 0.4\times {10}^{-3}}{1.5-1}$
$=\frac{240-480}{0.5}=-\frac{240}{0.5}$
or $W=-480J$