Question

A sample of an ideal gas is taken through a cycle a shown in 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 gas at $A$ is $1500J$, the internal energy at $C$ would be,

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

Answer 1

The correct option is C $1590J$

Given,
$Q_{AB}=50J$
$W_{BC}=-40J$
$U_A=1500J$
$Q_{BC}=0J$
For $AB$ process,
$\Delta W_{AB}=0$ as $V=\text{constant}$
From the first law of thermodynamics,
$Q=\Delta U+W$
$\therefore Q_{AB}=\Delta U_{AB}=50J$
$U_A=1500J$
$\therefore U_B=(1500+50)J=1550J$

For $BC$ process,
According to first law of thermodynamics,
$Q=\Delta U+W$
$Q_{BC}=0J$
$W_{BC}=-\Delta U_{BC}=-40J$
$\therefore \Delta U_{BC}=40J$
$\therefore U_C=(1550+40J)=1590J$

Final answer: (𝑎)