Answer to Question #87601 in Molecular Physics | Thermodynamics for Lesly

Question

An ideal gas initially is allowed to expand isothermally until its volume is 0.75 L and its pressure is 8 kPa. Then it is heated at constant volume until its pressure is 1 kPa.
1. Calculate the work done by the gas. Answer in units of kJ.
2. Find the heat added to the gas during this process. Answer in units of kJ.

Answer 1

First of all, I would like to note that the task contains either a typos or an error and is incomplete. Let me explain this in details.

1) the 2nd stage of the process is isochoric heating, thus the pressure should increase in comparison to the initial value but in the task we have 1 kPa against 8 kPa that is wrong, i.e., it should be:

$$Q = \Delta U = \frac{3}{2} V \Delta p \\ Q > 0 \, \Rightarrow \, \Delta p > 0$$

2) the 1st stage of the process is isothermal expansion. However, there is absolutely no information on any initial parameter (either volume, or pressure, or even temperature). Hence, if we consider the PV-diagram of the process, for instance, there is infinitely big number of states that evolve to the present state with known (P₂,V₂) values (given in the tast), i.e.

$$pV = \nu R T$$

even for a fixed T has infinitely big number of states satisfying

$$p_1 V_1 = p_2 V_2,$$

where we know only (p₂,V₂) values and nothing on (p₁,V₁)

So, I will give the general solution for such a process without substituting the numbers.

Q1. The work is done by gas only during the 1st stage - isothermal expansion (because there is no change in volume during the 2nd stage - isochoric heating). The corresponding expression is

$$A_{12} = \nu R T \ln{\frac{V_2}{V_1}} = p_2 V_2 \ln{\frac{V_2}{V_1}}$$

Thus we need either the V₁ value, or p₁ value, so that V₁ can be obtained as follows:

$$V_1 = V_2 \frac{p_2}{p_1}$$

Q2. The heat added to a gas can be calculated as follows:

$$Q=Q_{12} + Q_{23},$$

where

$$Q_{12} = A_{12} = p_2 V_2 \ln{\frac{V_2}{V_1}} \quad (isothermal \, process)\\ Q_{23} = \Delta U_{23} = \frac{3}{2} \nu R (T_3 - T) = \frac{3}{2} (p_3 - p_2) V_2 \quad (isochoric \, process)$$

Finally,

$$Q = p_2 V_2 \ln{\frac{V_2}{V_1}} + \frac{3}{2} (p_3 - p_2) V_2$$

In order to derive both answers in J (or kJ), one should take into account during the substitution that

$$1 \, L = 10^{-3} \, m^3\\ 1 \, kPa = 10^3 \, Pa\\ 1 \, kJ = 10^3 \, J$$