Question

A fixed mass of a gas is first heated isobarically to double the volume and then cooled isochorically to decrease the temperature back to the initial value. By what factor would the work done by the gas decrease, had the process been isothermal?

• A. 2
• B. 1/2
• C. ln 2
• D. ln 3

Answer 1

The correct option is C ln 2

let initial volume, pressure and temperature of gas be $P_0$,$V_0$ and $T_0$
in first step gas undergoes in isobaric process so pressure will be constant during the process
so Pressure = $P_0$ and Volume =$2V_0$ and from $PV=nRT$ ;
Temperature = $2T_0$.
in next step gas undergoes in isochoric process so volume will be constant during this process
so Volume = $2V_0$, Temperature = $T_0$ and from $PV=nRT$
equation pressure will be = $P_0/2$. Work done in isochoric process is zero.
Work done in isobaric and isochoric process is $W=nR{T}_o$

so if we see initial and final temperature values are same so lets consider it had been a isothermal process so $\Delta W=nR{T}_{0}ln\frac{V_2}{V_1}=nR{T}_{o}ln2$

So work done decreases by a factor of $ln2$.