Question

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from \( P_1 \) to \( P_2 \) is \( \alpha \) Joule \((P_1 = 21.7 \text{ Pa}, P_2 = 30 \text{ Pa}, C_v = 21 \text{ J/K mol}, R = 8.3 \text{ J/mol K})\). The value of \( \alpha \) is ________.

• A. 15
• B. 21
• C. 28
• D. 24

Hint:

For constant volume processes, heat supplied equals change in internal energy.

Answer 1

The Correct Option is B

To solve for the heat involved in the process from \( P_1 \) to \( P_2 \), we'll use the concept of an isochoric process, where volume remains constant. In this process, the heat added or removed is given by:

\(Q = nC_v \Delta T\)

where:

• \(n\) is the number of moles of the gas,
• \(C_v\) is the molar heat capacity at constant volume,
• \(\Delta T\) is the change in temperature.

First, we know:

• \(n = 10 \text{ moles}\)
• \(C_v = 21 \text{ J/K mol}\)
• \(P_1 = 21.7 \text{ Pa}\)
• \(P_2 = 30 \text{ Pa}\)

The ideal gas law states:

\(PV = nRT\)

For an isochoric process \( V \) and \( n \) are constant, thus:

\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)

This can be rearranged to find the change in temperature:

\(\frac{T_2}{T_1} = \frac{P_2}{P_1}\)

\(\Delta T = T_2 - T_1 = T_1\left(\frac{P_2}{P_1} - 1 \right)\)

Since \(P_1 = 21.7 \, \text{Pa}\) and \(P_2 = 30 \, \text{Pa}\), substituting these gives:

\(\Delta T = T_1\left(\frac{30}{21.7} - 1 \right)\)

We substitute this \(\Delta T\) into the heat equation:

\(Q = nC_v T_1\left(\frac{P_2}{P_1} - 1 \right)\)

\(Q = 10 \times 21 \times \frac{8.3}{R}\left(\frac{30}{21.7} - 1 \right)\)

Solving the above expression, determine \(Q\) as follows:

\(Q = 10 \times 21 \times \left(\frac{30}{21.7} - 1\right) = 10 \times 21 \times \left(\frac{30 - 21.7}{21.7}\right) = 10 \times 21 \times \frac{8.3}{21.7}\)

On calculating, \( Q = 10 \times 21 \times 0.382 = 21 \text{ J}\).

Therefore, the heat involved in the process is 21 Joules. The correct answer is: \(21\).

Thus, the correct option is 21.