## Essential University Physics: Volume 1 (3rd Edition)

(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)