First slide

Heat given

Question

One mole of Argon undergoes a process given by PV^3/2 = constant. If heat obtained by gas is Q and molar specific heat of gas in the process is C then which of the following is correct if temperature of gas changes by –26K (assume Argon as an ideal gas)

Moderate

A

C = 0.5R, Q = 13R
B

C = -0.5R, Q = 1.3R
C

C = -0.5R, Q = 13R
D

C = 0, Q = 13R

Solution

$\large C\, = \,{C_v} + \frac{R}{{1 - x}};\,where\,\,x\, = \,\frac{3}{2},{C_v} = \,\frac{3}{2}R$
$\large \therefore \,c\, = \,\frac{3}{2}R\, + \frac{R}{{\left( {1 - \frac{3}{2}} \right)}}\, = \, - \frac{R}{2}$
Also, $\large Q\, = \,n.c.\Delta T\,\, = \,\,1.\left( { - \frac{R}{2}} \right)\left( { - 26} \right)$
= 13R

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