Answer:
(a) Refer part (d)
of Q. No. 12.34
(b) \[Q=d{{U}_{BC}}+d{{W}_{BC}}\]
Now
\[d{{V}_{BC}}=\,{{C}_{\upsilon }}\,({{T}_{C}}-{{T}_{B}})=\,\frac{3}{2}\,R\,({{T}_{C}}\,-{{T}_{B}})\]
\[=\frac{3}{2}\,(R{{T}_{C}}-\,R{{T}_{B}})\,=\frac{3}{2}\,\,({{P}_{B}}{{V}_{C}}-{{P}_{B}}{{V}_{B}})\]
\[=\frac{3}{2}\,{{P}_{B}}\,({{V}_{C}}-{{V}_{B}})\]
and
\[d{{W}_{BC}}=PB\text{ }({{V}_{C}}-{{V}_{B}})\]
\[\therefore
\] \[Q=\frac{5}{2}\,{{P}_{B}}\,({{V}_{C}}-{{V}_{B}})\]
(c)
\[{{Q}_{CD}}=0\] (as no heat is exchanged between system and its surrounding
during adiabatic process)
(d)
Proceed as in (b)
\[Q=\,\frac{5}{2}\,{{P}_{A}}\,({{V}_{A}}-{{V}_{D}})\]
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