Question

A perfect gas goes from state $A$ to state $B$ by absorbing $8\times {10}^5\text{J}$ of heat and doing $6.5\times {10}^5\text{J}$ of external work. It is now transferred between the same two states in another process in which it absorbs ${10}^5\text{J}$ of heat. In the second process,

• A. Work done on gas is ${10}^5\text{J}$
• B. Work done on gas is $0.5\times {10}^5\text{J}$
• C. Work done by gas is $0.5\times {10}^5\text{J}$
• D. Work done by gas is ${10}^5\text{J}$

Answer 1

The correct option is B Work done on gas is $0.5\times {10}^5\text{J}$

Given,
Heat absorbed through process $1$ $\Delta Q)=8\times {10}^5\text{J}$
Work done by the system $\left(\Delta W\right)=6.5\times {10}^5\text{J}$
Heat absorbed through process $2$ $\left(\Delta Q\right)={10}^5\text{J}$
Since, system is transferred between same two states A and B we can deduce, $\Delta U$ remains same.
$\therefore$ From 1st law
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta U=\Delta Q-\Delta W=8\times {10}^5-6.5\times {10}^5$
$\Delta U=1.5\times {10}^5\text{J}$
For another process,
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta W=\Delta Q-\Delta U={10}^5-(1.5\times {10}^5\text{J})$
$\Delta W=-0.5\times {10}^5\text{J}$