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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Two moles of a diatomic ideal gas is taken through the process $PT$ = const. Its temperature is increased from $T_oK$ to $2T_oK$. Find the work done by the system.

$\begin {array} {1 1} (A)\;RT_o & \quad (B)\;4RT_o \\ (C)\;2RT_o & \quad (D)\;3RT_o \end {array}$

1 Answer

$W = \int P\: dV$
Here, $PT = P_1T_1 = P_2T_2 = c(constant)$
$ PT = c;$
$ \Rightarrow P. PV/nR = c$
$P^2V = ncR $
$ \Rightarrow P = \sqrt{ \bigg( \large\frac{ncR}{V} \bigg)}$
So, $ \int P\: dV = \sqrt{ncR}\:  [2(\sqrt{V_2} - \sqrt{V_1} )]$
$= 2[\sqrt{(nRP_2T_2V_2)} - \sqrt{(nRP_1T_1V_1)}] = 2nR(T_2 – T_1) = 4RT_o$
Ans : (B)
answered Mar 15, 2014 by thanvigandhi_1
edited Mar 21, 2014 by balaji.thirumalai
 

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