Question

An ideal monoatomic gas is carried along the cycle $A B C D A$ as shown in the figure. The total heat absorbed during this process is

• A. $10.5 \,p _{0} \,V _{0}$
• B. $7.5\, p _{0} \,V _{0}$
• C. $2.5 \,p_{0} \,V_{0}$
• D. $1.5 \,p _{0} \,V _{0}$

Answer 1

Answer: (A)

Heat absorbed means heat is extracted from source, i.e.
$Q$ must be positive.
This occurs along path
$A \rightarrow B \rightarrow C$
$\therefore $ Heat added $=\Delta Q_{A B C}=\Delta U_{A B C}+\Delta W_{A B C}$
$=n C_{V}\left(T_{C}-T_{A}\right)+3 p_{0}\,\left(2 V_{0}-V_{0}\right)$
$=n \frac{3}{2} R\left(T_{C}-T_{A}\right)+3 p_{0}$
$V_{0}=\frac{3}{2}\left(n R T_{C}-n R T_{A}\right)+3 p_{0} V_{0}$
$=\frac{3}{2}\left(p_{C} \,V_{C}-p_{A} \,V_{A}\right)+3 p_{0} \,V_{0}$
$=\frac{3}{2}\left(3 p_{0} \,2 V_{0}-p_{0} \,V_{0}\right)+3 p_{0}\, V_{0}$
$=\frac{3}{2} \times 5 p_{0} \,V_{0}+3 p_{0}$
$V_{0}=\frac{21}{2} p_{0}\, V_{0}=10.5 p_{0}\, V_{0}$