Answer
Please see the work below.
Work Step by Step
(a) As the process is isothermal so $\Delta E_{int}=0$ and $T_2=T_1$
Now $W=-Q$
$W=-1.75KJ$ (Negative sign shows that the work is done by gas)
$T_{final}=255K$ (as temperature is constant)
(b) As the volume is constant so $W=0$
$\implies T_{final}=T_1+\frac{Q}{nC_V}$
We plug in the known values to obtain:
$T_{final}=255+\frac{1.75\times 10^3}{35\times \frac{5}{2}\times 8.314}$
$T_f=279K$
(c) We know that
$T_{final}=T_{initial}+\frac{Q}{nC_P}$
We plug in the known values to obtain:
$T_{final}=255+\frac{1.75\times 10^3}{35\times \frac{7}{2}8.314}$
$T_{final}=272K$
Now $W=-nR(T_2-T_1)$
We plug in the known values to obtain:
$W=-35\times 8.314\times (272-255)=-495 J$ (negative sign shows that work is done b by gas)
