I can't take this Heat!

It's a polytropic process with n = -1.

As kx = PA and V is directly proportional to x and hence $P {V}^{-1} = constant$

The compression in the spring at an instant can be given as $x=\frac { V }{ A } \\ { x }_{ 0 }=\frac { { V }_{ 0 } }{ A }$

Lets's assume the process is a quasi-equilibrium process. Therefore the pressure in the gas at any instant is the pressure exerted by the spring ${ P }.A=k(\frac { V }{ A } )$ . Now from gas equation $PV=mRT\\ (k\frac { V }{ { A }^{ 2 } } )V=mRT\\ { V }^{ 2 }=\frac { mR{ A }^{ 2 } }{ k } T$

Work done by gas = work done by spring $\frac { 1 }{ 2 } k({ x }^{ 2 }-{ { x }_{ 0 } }^{ 2 })=\frac { 1 }{ 2 } k({ \frac { { V }^{ 2 } }{ {A}^{2} } }-\frac { { { V }_{ 0 } }^{ 2 } }{ { A }^{ 2 } } )\\ \qquad \qquad \qquad =\frac { 1 }{ 2 } k({ \frac { \frac { mR{ A }^{ 2 } }{ k } T }{ { A }^{ 2 } } }-\frac { { { V }_{ 0 } }^{ 2 } }{ { A }^{ 2 } } )\\ \qquad \qquad \qquad =\frac { 1 }{ 2 } mRT\quad -\quad \frac { 1 }{ 2 } \frac { k{ { V }_{ 0 } }^{ 2 } }{ { A }^{ 2 } }$

From 1st law : $\triangle U={ Q }_{ in }-{ W }_{ out }\\ { Q }_{ in }=\triangle U+{ W }_{ out }\\ { Q }_{ in }=\frac { 3 }{ 2 } mR(T-{ T }_{ 0 })\quad +\quad (\frac { 1 }{ 2 } mRT\quad -\quad \frac { 1 }{ 2 } \frac { k{ { V }_{ 0 } }^{ 2 } }{ { A }^{ 2 } } )\\ { Q }_{ in }=\frac { 3 }{ 2 } mRT\quad +\quad \frac { 1 }{ 2 } mRT\quad +\quad constant$

$Heat\quad capacity=\frac { d{ Q }_{ in } }{ dT } =\frac { 3 }{ 2 } mR+\frac { 1 }{ 2 } mR=\quad 2mR\\ \qquad \qquad \qquad \qquad \qquad \quad \quad =\quad 2\frac { { { P } }_{ 0 }{ V }_{ 0 } }{ { T }_{ 0 } } (in\quad S.I\quad units)\\ \qquad \qquad \qquad \qquad \qquad \quad \quad =\quad 2.128\quad J/K$