Answer
Please see the work below.
Work Step by Step
(a) We know that
$T=\frac{PV}{nR}$
$\implies T_A=\frac{(150\times 10^3KPa)(1.00m^3)}{(67.5mol)(8.31J/mol.K)}$
$T_A=267K$
At point B:
$T_B=\frac{(50\times 10^3KPa)(4.00m^3)}{(67.5mol)(8.31J/mol.K)}=357K$
At point C:
$T_C=\frac{(50\times 10^3KPa)(1.00m^3)}{(67.5mol)(8.31J/mol.K)}=89.1K$
(b) We know that for process $A\rightarrow B$, the temperature rises and the gas does the work; therefore, we know that heat enters the system.
For process $B\rightarrow C$, the temperature drops and work is done on the gas, hence heat leaves the system.
For process $C\rightarrow A$, the temperature rises and no work is done on or by the gas, hence heat enters the system.
(c) We know that for process $A\rightarrow B$
$Q=\frac{3}{2}nR\Delta T+\frac{1}{2}\Delta P\Delta V$
$\implies Q=\frac{3}{2}(67.5mol)(8.31J/mol.K)(357K-268K)+\frac{1}{2}(150KPa+50KPa)(4-1)m^3=376KJ$
For process $B\rightarrow C$
$Q=\frac{3}{2}nR\Delta T+\frac{1}{2}P\Delta V$
$\implies Q=\frac{3}{2}(67.5mol)(8.31J/mol.K)(8.91K-357K)+\frac{1}{2}(1-4)m^3=-375KJ$
For process $C\rightarrow A$
$Q=\frac{3}{2}nR\Delta T+\frac{1}{2}P\Delta V$
$\implies Q=\frac{3}{2}(67.5mol)(8.31J/mol.K)(267K-89.1K)+0$
$\implies Q=150KJ$
