# 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

"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.

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 $$W=\frac{nRT_1}{V_2}(V_3-V_2)=nRT_1\left(\frac{V_3}{V_2}-1\right)=nRT_1$$

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 $$W_{s23}=nRT_1$$