Question

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

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

Answer 1

Answer: (B)

From first law of thermodynamics
$d U=d Q-d W $
$d U=8 \times 10^{5}-6.5 \times 10^{5} $
$d U=1.5 \times 10^{5} J$
In the second process, $d U$ remains the same.
$\therefore d W=d Q-d U=1 \times 10^{5}-1.5 \times 10^{5}$
$d W=-0.5 \times 10^{5} J$
Negative sign shows that work is done on the gas.