A monatomic ideal gas at 27°C undergoes an isobaric process from state A to B, followed by an isochoric process from state B to C. What is the total work done by the gas in these two processes? (A = 2 atm, 1 L; B = 2 atm, 2 L; C = 1 atm, 2 L)
A) 203 J
B) 231 J
C) 187 J
D) 195 J

Question

Quizplus · Accepted Answer

The total work done by the gas in these two processes can be calculated by considering the work done in each process separately. 1. Isobaric process from A to B: In an isobaric process, the work done by the gas is given by $W = P \Delta V$, where $P$ is the pressure and $\Delta V$ is the change in volume. From state A to B, the pressure is constant at 2 atm (which is equivalent to $2 \times 101.325 = 202.65 \, \text{kPa}$) and the volume changes from 1 L to 2 L, so $\Delta V = 1 \, \text{L} = 0.001 \, \text{m}^3$. Therefore, the work done is $W = 202.65 \times 0.001 = 202.65 \, \text{J}$.2. Isochoric process from B to C: In an isochoric process (constant volume), the volume does not change, which means no work is done by the gas ($W = 0$) because work is a product of pressure change and volume change, and here the volume change is zero.Adding the work done in both processes gives a total work of $202.65 \, \text{J} + 0 \, \text{J} = 202.65 \, \text{J}$, which is approximately 203 J when rounded to the nearest whole number.

A Monatomic Ideal Gas at 27°C Undergoes an Isobaric Process