Question

A monatomic ideal gas has pressure p1 and temperature T1. It is contained in a cylinder of volume V1 with a movable piston, so that it can do work on the outside world. Consider the following three-step transformation of the gas: The gas is heated at constant volume until the pressure reaches Ap1 (where A>1). The gas is then expanded at constant temperature until the pressure returns to p1. The gas is then cooled at constant pressure until the volume has returned to V1.
It may be helpful to sketch this process on the pVplane.
How much heat Q1 is added to the gas during step 1 of the process?
Express the heat added in terms of p1, V1, and A.
How much work W2 is done by the gas during step 2?
Express the work done in terms of p1, V1, and A.
How much work W3 is done by the gas during step 3?
If you've drawn a graph of the process, you won't need to calculate an integral to answer this question.
Express the work done in terms of p1, V1, and A.

Answer 1

1) $Q_1=\Delta U=\frac{3}{2}P_1 V_1 (A-1)$

The heat absorbed by the gas can be found by using the 1st law of thermodynamics:

$\Delta U=Q-W$

where

$\Delta U$ is the change in internal energy

Q is the heat absorbed

W is the work done

Since the first process occurs at constant volume, the work done is zero:

$W=\int pdV = 0$

So the equation becomes

$\Delta U=Q$

The change in internal energy is given by

$\Delta U=n c_v (T_2-T_1)$ (1)

where

n is the number of moles of the gas, which can be found by using the ideal gas law with the initial conditions of the gas:

$n=\frac{p_1 V_1}{RT_1}$ (2)

where R is the gas constant,

$c_v =\frac{3}{2}R$ is the specific heat at constant volume

$T_2-T_1$ is the change in temperature. The temperature T2 can calculate again by using the ideal gas law at the new conditions of the gas, after its pressure has reached Ap1:

$T_2 = \frac{AP_1 V_1}{nR}$ (3)

Combining (1), (2) and (3), we find:

$Q=\Delta U=\frac{3}{2}P_1 V_1 (A-1)$

2) $W_2=AP_1 V_1 ln(A)$

The work done during an isothermal process is given by:

$W=nRTln(\frac{V_f}{V_i})$ (4)

where in this case we have:

$n=\frac{p_1 V_1}{RT_1}$  (number of moles)

$T=T_2= \frac{AP_1 V_1}{nR}$ is the constant temperature of the process, found in the previous part

$V_f = V_2$ is the final volume, which  can be found again by using the ideal gas law:

$V_2 = \frac{nRT_2}{P_1}$

$V_i=V_1$ is the initial volume

Substituting all the quantities into (4), we find:

$W=AP_1 V_1 ln(A)$

3) $W_3=P_1 V_1 (1-A_1)$

Step 3 is an isobatic process (constant pressure), so the work done can be simply calculated as

$W=p (V_f-V_i)$

where:

$p=P_1$ is the constant pressure of the process

$V_f = V_1$ is the final volume

$V_i=V_2 = \frac{nRT_2}{P_1}$ is the initial volume

Substituting into the equation, and keeping in mind that

$n=\frac{p_1 V_1}{RT_1}$

$T_2= \frac{AP_1 V_1}{nR}$

We find:

$W=P_1 V_1 (1-A_1)$