Answer
(a) For process (3) $530KJ$; For process (4) $477KJ$
(b) $212KJ$
(c) $795KJ$
Work Step by Step
(a) We can find the required heat added to the gas during the two processes as follows:
For process(3),initial temperature is given as
$T_i=\frac{P_iV_i}{nR}$
$\implies T_i=\frac{(106KPa)(1m^3)}{(60mol)(813J/mol.K)}$
$T_i=212.6K$
For process(3), final temperature is given as
$T_f=\frac{P_iV_f}{nR}$
$\implies T_f=\frac{(106\times 10^3Pa)(1m^3)}{(60mol)(8.31J/mol.K)}$
$T_f=637.8K$
Now the heat added for process(3) is given as
$Q_p=\frac{5}{2}nR(T_f-T_i)$
We plug in the known values to obtain:
$Q_p=\frac{5}{2}(60mol)(8.31J/mol.K)(637.8K-212.6K)$
$Q_p=530KJ$
For process (4), the final temperature is given as
$T_f=\frac{P_fV_f}{nR}$
$\implies T_f=\frac{(212KPa)(3m^3)}{(60mol)(8.31J/mol.K)}$
$T_f=1275.57K$
Now $Q_v=\frac{3}{2}nR(T_f-T_i)$
We plug in the known values to obtain:
$Q_v=\frac{3}{2}(60mol)(8.31J/mol.K)(1275.57K-637.8K)$
$Q_v=477KJ$
(b) The work done can be determined as
$W=P\Delta V$
We plug in the known values to obtain:
$W=(106\times 10^3Pa)(3m^3-1m^3)$
$W=212KJ$
(c) We know that
$\Delta U=Q_p+Q_v-W$
We plug in the known values to obtain:
$\Delta U=530KJ+477KJ-212KJ$
$\Delta U=795KJ$
