Question

A perfect gas goes from a state $A$ to state $B$ by absorbing $8\times {10}^5\text{J}$ and by doing $6.5\times {10}^5\text{J}$ of external work. It is taken from same initial state $A$ to final state $B$ in another process in which it absorbs ${10}^5\text{J}$ of heat, then work done in the second process

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

Answer 1

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

Heat is absorbed by system {gas}, $\Delta Q=+8\times {10}^5\text{J}$
work done by system, $\Delta W=+6.5\times {10}^5\text{J}$
According to the $1^{\text{st}}$ law of thermodynamics
$\Delta Q=\Delta U+\Delta W$
$8\times {10}^5=\Delta U+6.5\times {10}^5$
$\Rightarrow \Delta U=+1.5\times {10}^5\text{J}........\left(1\right)$
For the second process,
Heat absorbed by the system $\Delta Q^{\prime }=+{10}^5\text{J}$
Again using First law,$\Delta Q=\Delta U+\Delta W$
We write that,
$\Delta Q^{\prime }=\Delta U+\Delta W^{\prime }$
$\Delta U$ will stay the same as initial and final states of gas is same.
Now, From the data given in the question and $\left(1\right)$
${10}^5=1.5\times {10}^5+\Delta W^{\prime }$
$\Delta W^{\prime }=-0.5\times {10}^5\text{J}$
(negative sign indicates work is being done on the gas).
Thus, option (b) is the correct answer.