Question

An ideal monoatomic gas is taken through cycle $A\to B\to C\to A$. What is the work done by the gas in the step $B\to C$?

• A. $\frac{4P_0{V}_0}{2}-0-8P_0{V}_0=-6P_0{V}_0$
• B. $-6P_0{V}_0$
• C. $9P_0{V}_0$
• D. $2P_0{V}_0$

Answer 1

The correct option is A $\frac{4P_0{V}_0}{2}-0-8P_0{V}_0=-6P_0{V}_0$

An ideal monoatomic gas is taken through cycle $A\to B\to C\to A$. The work done by
the gas in the step $B\to C$ is $\frac{4P_0{V}_0}{2}-0-8P_0{V}_0=-6P_0{V}_0$.
Note: The work done by the gas in the step $A\to B$ is zero.
The work done by the gas during the cycle $A\to B\to C\to A$ is $\frac{\Delta P\times \Delta V}{2}=\frac{(2P_0-P_0)(6V_0-2V_0)}{2}=\frac{4P_0{V}_0}{2}$.
The work done by the gas in the step $B\to B$ is
$W=P\Delta V=2P_0\times (6V_0-2V_0)=8P_0{V}_0$
The work done by
the gas in the step $B\to C$ is
$\frac{4P_0{V}_0}{2}-0-8P_0{V}_0=-6P_0{V}_0$