(200)

$w=-P_{\text{ext}}\Delta V=-P_{\text{ext}}(V_2-V_1)=-5(20-60) \text{atm L}=200 \text{atm L}$

By first law of thermodynamics:

$\Delta V=-w+q$

For isothermal process, $\Delta U=0,$ so $q=-w=-200 \text{atm L}$

(274)

Work done (W) by gas against a constant pressure is given by:

$W=-P_{\text{ext}}(V_f-V_i)=-1(2V-V)=-V$                      ...(i)

Change in internal energy ($∆U$) for an ideal gas undergoing any process is:

$\Delta U=n{C}_v(T_f-T_i)=n\times \frac{5}{2}R(T_f-298)$                      ...(ii)

By first law of thermodynamics:

$\Delta U=q+W$

Since q = 0 for an adiabatic process,

$\Delta U=W$

Put $∆U$ from (ii) and W from (i)

$n\times \frac{5}{2}R(T_f-298)=-V$                       ...(iii)

Apply ideal gas equation at initial conditions,

$P_i{V}_i=nR{T}_i$

$5V=nR\times 298$

$V=\frac{nR\times 298}{5}$                                    ...(iv)

Put V from (iv) in (iii)

$n\times \frac{5}{2}R(T_f-298)=-\frac{nR\times 298}{5}$

$\frac{5}{2}(T_f-298)=-\frac{298}{5}$

$T_f-298=-\frac{298\times 2}{5\times 5}$

$T_f=274.16\approx 274 K$