Question

A cylinder of ideal gas is closed by an $8kg$ movable piston (area $60c{m}^2$). Atmospheric pressure is ${10}^{5}Pa$. When the gas is heated from $30°C$ to $100°C$, the piston rises by $20cm$. The piston is then fixed in its place and the gas is cooled back to $30°C$. Let $\Delta Q_1$ be the heat added to the gas in the heating process and $\left|\Delta Q_2\right|$ the heat lost during cooling. Then the value of $\Delta Q_1-\left|\Delta Q_2\right|$ will be

• A. $0$
• B. $136J$
• C. $-136J$
• D. $-68J$

Answer 1

The correct option is B $136J$


In case 1: Temperature from ${30}^{o}C\to {100}^{o}C$
$\Delta Q_1=\Delta U_1+\Delta W_1\Delta Q_1=\Delta U_1+P\Delta V....\left(1\right)$

Total pressure at piston


$P=P_0+\frac{mg}{A}$
Using the given data,
$P={10}^5+\frac{8\times 10}{60\times {10}^{-4}}=1.13\times {10}^{5}Pa$

Putting this value in eq.(1)

$\Delta Q_{!}=\Delta U_1+(1.13\times {10}^5)(60\times {10}^{-4}\times 0.2)=\Delta U_1+136$

Case 2: ${100}^{o}C\to {30}^{o}C$
$\Delta Q_2=\Delta U_2+\Delta W_2$
$\Delta W_2=0$ (given)
$\Delta Q_2=\Delta U_2=-\Delta U_1$
(because $\Delta U$ depends only on initial and final state)
$\left|\Delta Q_2\right|=\Delta U_1$

So,
$\Delta Q_1-\left|\Delta Q_2\right|=\Delta U_1+136-\Delta U_1\Delta Q_1-\left|\Delta Q_2\right|=136J$