# Work done by isothermal gas expansion followed by an isobaric expansion

A mono-atomic ideal gas of $2$ moles undergoes an isothermal expansion which makes its volume double. Then the volume of the gas is doubled again through an isobaric process. Given a starting temperature of $350$ Kelvin:

• how much work is done by the gas during the isobaric expansion?
• how much heat is transferred during the isothermal expansion?

note: R = $8.31$ J/(mol*K) and Cp = $5/2$ * R

The problem gives me this data:

• $n$ = 2 mol
• $T_1$ = $350$ K
• $R$ = $8.31$ J/(mol*K)
• $C_p$ = $\frac5 2$ $R$
• $V_3 = 2V_2 = 4V_1$

This is the pressure-volume diagram: The formula that I would use to find the work for the isobaric expansion would be $W = nR(T_3 - T_2)$

But I don't know how to find the temperature at the end of the expansion, because the problem doesn't give me pressure or volume. So I can't use $P_iV_i = nRT_i$ to start finding it. I don't know if I'm missing something very important or the problem is wrong

Actually, there is enough information to solve this. $$W=P_2(V_3-V_2)$$together with $$P_2=\frac{nRT_2}{V_2}=\frac{nRT_1}{V_2}$$So $$W=\frac{nRT_1}{V_2}(V_3-V_2)=nRT_1\left(\frac{V_3}{V_2}-1\right)=nRT_1$$

"I'm missing something very important or the problem is wrong". There isn't enough information to find numerical values for $p_1$ or $V_1$ separately. But all the other pressures and volumes can easily be expressed in terms of $V_1$. Leave $V_1$ in your workings out, and you'll find it cancels out, so you can give numerical answers to the questions.

The question is already perfectly answered, but I thought i'd add a little bit starting with the first law as I personally found it very helpful to get some sort of intuition for problems of this kind:

$$dU=dQ+W_s$$

From State 1 to State 2:

Since $$dU=c_v * dT$$ and the process is isothermal, $dU=0$ and so $dQ = - W_s$

Now by using the ideal gas law:

$$W_{s12}= -\int pdV = -nRT_1*ln\frac{2V_1}{ V_1}=-nRT_1*ln(2)$$

State 2 to State 3:

Since $p*V = const$, the pressure of State 2 is

$$p_2=\frac{p_1*V_1}{2V_1} =\frac{p_1}{2}$$

the process is isobaric, so the shaft work simplifies to

$$W_{s23}= -\int pdV = p_2*\Delta V = \frac{p_1}{2} * (4V_1-2V_1) = p_1*V_1$$

So in terms of known quantities $$W_{s23}=nRT_1$$