JEE Chemistry 2023 Question with Solution

Given: The process involves three steps:

• $1 \rightarrow 2$: isobaric process
• $2 \rightarrow 3$: isochoric process
• $3 \rightarrow 1$: isothermal process

Find: The magnitude of the total work done.

The total work done is

$$W = W_{1 \rightarrow 2} + W_{2 \rightarrow 3} + W_{3 \rightarrow 1}$$

For each step,

$$W_{1 \rightarrow 2} = -P(V_2 - V_1)$$ $$W_{2 \rightarrow 3} = 0$$

for the isochoric process, and

$$W_{3 \rightarrow 1} = -P_1 V_1 \ln\left(\frac{V_2}{V_1}\right)$$

Substituting the values,

$$W = \left[-1 \times (40 - 20) + 0\right] + \left[-1 \times 20 \ln\left(\frac{20}{40}\right)\right]$$ $$W = -20 + 20 \ln 2$$

Using $\ln 2 = 0.3$,

$$W = -20 + 20 \times 2.3 \times 0.3$$ $$W = -20 + 6.2$$ $$W = -6.2 \, \text{bar L}$$

Therefore, the magnitude of work is $6.2 \, \text{bar L}$.

Now convert to joules using $1 \, \text{bar L} = 100 \, \text{J}$:

$$|W| = 6.2 \, \text{bar L} \times 100 = 620 \, \text{J}$$

Therefore, the required numerical answer is 620.

Given: A cyclic path with one isobaric, one isochoric, and one isothermal step.

Find: The net work magnitude.

Evaluate work segment by segment:

1. In the isobaric step, work equals pressure multiplied by change in volume.
2. In the isochoric step, volume does not change, so work is zero.
3. In the isothermal step, use the logarithmic expression for work.

From the provided working,

$$W_{1 \rightarrow 2} = -1 \times (40 - 20) = -20 \, \text{bar L}$$ $$W_{2 \rightarrow 3} = 0$$ $$W_{3 \rightarrow 1} = -1 \times 20 \ln\left(\frac{20}{40}\right) = 20 \ln 2$$

Adding them,

$$W = -20 + 20 \ln 2$$

The extracted solution then uses the numerical value to obtain

$$W = -6.2 \, \text{bar L}$$

Hence,

$$|W| = 6.2 \, \text{bar L} = 620 \, \text{J}$$

So the final answer is 620.