Question

A sample of an ideal gas is taken through a cycle as shown in the figure. It absorbs $50J$ of energy during the process AB, no heat during BC, rejects $70J$ during CA. $40J$ of work is done on the gas during BC. the internal energy of a gas at A is $1500J$, the internal energy at C would be

• A. $1590J$
• B. $1620J$
• C. $1540J$
• D. $1570J$

Answer 1

Answer: (A)

$1590J$

$Given,Q_{AB}=50J{Q}_{BC}=0J{Q}_{CA}=-70J{W}_{BC}=-40J{U}_A=1500JForprocessA\to B,\Delta V_{AB}=0⟹\Delta W_{AB}=0W_{AC}=W_{AB}+W_{BC}=0+(-40)=-40J{Q}_{AC}=Q_{AB}+Q_{BC}=50+0=50JUsingFirstlawofThermodynamics,\Delta U=\Delta Q-\Delta WTherefore,U_C-U_A=U_{AC}=Q_{AC}-W_{AC}\Rightarrow U_C=U_A+Q_{AC}-W_{AC}=1500+50-(-40)=1590JTheanswerisoption\left(A\right).$