Answer
(a) process 1: $159KJ$; process 2: $1060KJ$
(b) $424KJ$
(c) $795KJ$
Work Step by Step
(a) We can find the required heat added in the two processes as follows:
$T_i=\frac{P_iV_i}{nR}$
$\implies T_i=\frac{(106\times 10^3Pa)(1m^3)}{(60mol)(8.31J/mol.K)}$
$T_i=212.6K$
Similarly $T_f=\frac{P_fV_f}{nR}$
$\implies T_i=\frac{(212\times 10^3Pa)(1m^3)}{(60mol)(8.31J/mol.K)}$
$T_i=425.25K$
The heat added in process(1) is given as
$Q_v=\frac{3}{2}nR(T_f-Ti)$
We plug in the known values to obtain:
$Q_v=\frac{3}{2}(60mol)(8.31J/mol.K)(425.25K-212.6K)$
$Q_v=159KJ$
Now $T_f=\frac{P_fV_f}{nR}$
We plug in the known values to obtain:
$T_f=\frac{(212KPa)(3m^3)}{(60mol)(8.31J/mol.K)}$
$T_f=1275.6K$
The heat added in process (2) is given as
$Q_p=\frac{5}{2}nR(T_f-T_i)$
We plug in the known values to obtain:
$Q_p=\frac{5}{2}(60mol)(8.31J/mol.K)(1275.6K-425.25K)$
$Q_p=1060KJ$
(b) We can find the required work done as follows:
$W=P\Delta V$
We plug in the known values to obtain:
$W=(212\times 10^3Pa)(3m^3-1m^3)$
$W=424KJ$
(c) We know that
$\Delta U=Q_v+Q_p+W$
We plug in the known values to obtain:
$\Delta U=159KJ+1060KJ-424KJ$
$\Delta U=795KJ$
