Question

A diatomic gas ($\gamma = 1.4$) does 200 J of work when it is expanded isobarically. The heat given to the gas in the process is:

• A. 850 J
• B. 800 J
• C. 600 J
• D. 700 J

Answer 1

The Correct Option is D

To determine the heat supplied to the gas during an isobaric process, the first law of thermodynamics, \(Q = \Delta U + W\), is applied. In this equation, \(Q\) represents the heat exchanged, \(\Delta U\) signifies the change in internal energy, and \(W\) denotes the work done by the system.

We are given that the work done by the gas is \(W = 200 \text{ J}\) and our objective is to find \(Q\).

For a diatomic gas undergoing an isobaric process, the change in internal energy is related to the temperature change by \(\Delta U = nC_v\Delta T\), and the work done is given by \(W = nR\Delta T\).

The specific heat capacities, \(C_p\) and \(C_v\), are linked to the specific heat ratio \(\gamma = 1.4\). For a diatomic gas, these are:

• \(C_p = \frac{7}{2}R\)
• \(C_v = \frac{5}{2}R\)

The ratio of specific heats is:

• \(\gamma = \frac{C_p}{C_v} = \frac{1.4}{1}\)

The heat exchanged in an isobaric process can be expressed as \(Q = nC_p \Delta T\). Substituting the expressions for \(\Delta U\) and \(W\) into the first law gives \(Q = \Delta U + W\).

The relationship between heat and work, considering the specific heats, simplifies to \(Q = \frac{C_p}{C_v} \times W\) because \(\frac{nC_p\Delta T}{nR\Delta T} = \frac{C_p}{R}\) and \(\frac{\Delta U}{W} = \frac{nC_v\Delta T}{nR\Delta T} = \frac{C_v}{R}\). Thus \(Q = \Delta U + W = W(\frac{C_v}{R} + 1) = W(\frac{C_v+R}{R})\). Since \(C_p = C_v + R\), \(Q = W \frac{C_p}{R}\). Also \(W = nR\Delta T\), so \(Q = \frac{C_p}{R} W\).

Using the given values:

\(Q = \frac{C_p}{C_v} \times W = \frac{7/2 R}{5/2 R} \times W = \frac{7}{5} \times W\)

Substituting \(W = 200 \text{ J}\):

\(Q = \frac{7}{5} \times 200 \text{ J}\)

This calculation yields:

\(Q = 7 \times 40 \text{ J} = 280 \text{ J}\)

However, the relationship \(Q = nC_p \Delta T\) and \(W = nR \Delta T\) implies \(Q/W = C_p/R\). Thus \(Q = W \frac{C_p}{R} = 200 \text{ J} \times \frac{7/2 R}{R} = 200 \text{ J} \times \frac{7}{2} = 700 \text{ J}\).

Therefore, the heat transferred to the gas is 700 J.