Question

A gas is undergoing a cyclic process as shown in figure, where $P_0=2\text{bar}$ and $V_0=7{\text{m}}^3$. Find the workdone ($W$) by the gas and heat ($Q$) released by the gas in one cycle.

• A. $44\times {10}^5\text{J}$, $-44\times {10}^5\text{J}$
• B. $-44\times {10}^5\text{J}$, $44\times {10}^5\text{J}$
• C. $88\times {10}^5\text{J}$, $44\times {10}^5\text{J}$
• D. $-88\times {10}^5\text{J}$, $88\times {10}^5\text{J}$

Answer 1

The correct option is B $-44\times {10}^5\text{J}$, $44\times {10}^5\text{J}$

Given that,
$P_0=2\text{bar}$, $V_0=7{\text{m}}^3$
As we know that,
Work done $W=-$ (Area enclosed by $P-V$ graph)
[-ve sign since cycle is anticlockwise]

$W=-\pi R_1{R}_2$
$=-\pi \left(\frac{3P_0-P_0}{2}\right)\left(\frac{3V_0-V_0}{2}\right)$
$=-\pi P_0{V}_0$
$=-44\times {10}^5\text{J}$
So, work done by the gas is $-44\times {10}^5\text{J}$

In a thermodynamic cyclic process,
$\Sigma Q=\Sigma W$
So, in this process,
Net heat transfer will be $-44\times {10}^5\text{J}$.
That means, value of heat released in this process is $44\times {10}^5\text{J}$