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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

A gas in a piston cylinder assembly undergoes an expansion process where $PV^{1.5}$=constant.The initial pressure is 3 bar,the initial volume is $0.1m^3$ and the final volume is $0.2m^3$.Determine the work done for this process.

$\begin{array}{1 1}17.6kJ\\18.6kJ\\19.6kJ\\20.6kJ\end{array} $

1 Answer

Answer : 17.6kJ
$W_{1-2}=\int\limits_1^2 PdV$
$\qquad\;\;=C\large\frac{V_2^{-n+1}-V_1^{-n+1}}{-n+1}$
$\qquad\;\;=\large\frac{P_2V_2-P_1V_1}{1-n}$
The pressure at state 2 can be found using
$P_2V_2^n=P_1V_1^n$
$P_2=P_1\big(\large\frac{V_1}{V_2}\big)^n$
$\;\;\;\;\;=(3\times 10^5Pa)\big(\large\frac{0.1}{0.2}\big)^{1.5}$
$\;\;\;\;\;=1.06\times 10^5$Pa
Hence,$W=\big(\large\frac{(1.06\times 10^5Pa)(0.2m^3)-(3\times 10^5Pa)(0.1m^3)}{1-1.5}\big)$
$\Rightarrow 17.6kJ$
answered Sep 1, 2014 by sreemathi.v
 

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