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 find the heat given to the gas during the isobaric process, we need to use the first law of thermodynamics. The law states that:

\(Q = \Delta U + W\)

Where:

• \(Q\) is the heat exchanged with the system.
• \(\Delta U\) is the change in internal energy.
• \(W\) is the work done by the system.

Given that \(W = 200 \text{ J}\) as the work done by the gas and we need to find \(Q\).

For a diatomic gas, the relationship between the change in internal energy and the change in temperature during an isobaric process is given by:

\(\Delta U = nC_v\Delta T\)

and the work done is:

\(W = nR\Delta T\)

The specific heat capacities \((C_p \text{ and } C_v)\) are related to the specific heat ratio \(\gamma = 1.4\). For a diatomic gas:

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

The relationship between \(C_p\) and \(C_v\) is:

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

Rearranging the specific heat formula for an isobaric process, \(Q\) is:

\(Q = nC_p \Delta T = \Delta U + W\)

The formula naturally simplifies to:

\(\Delta U = nC_v\Delta T\)

Substituting values, we get:

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

Upon simplifying, we find:

\(Q = 280 \text{ J} \times 2.5 = 700 \text{ J}\)

Therefore, the heat given to the gas in this process is 700 J.

Answer 2

Given: - Work done by the gas during isobaric expansion: \(W = 200 \, \text{J}\) - For a diatomic gas, the ratio of specific heats \(\gamma = 1.4\).

Step 1: Relationship for an Isobaric Process

In an isobaric process, the heat supplied \(Q\) to the system is given by:

\[ Q = \Delta U + W \]

where \(\Delta U\) is the change in internal energy of the gas and \(W\) is the work done by the gas.

Step 2: Change in Internal Energy

The change in internal energy for a diatomic gas is given by:

\[ \Delta U = nC_V\Delta T \]

For a diatomic gas, the molar specific heat at constant volume \(C_V\) is:

\[ C_V = \frac{R}{\gamma - 1} = \frac{R}{1.4 - 1} = \frac{5R}{2} \]

The molar specific heat at constant pressure \(C_P\) is given by:

\[ C_P = C_V + R = \frac{5R}{2} + R = \frac{7R}{2} \]

Thus, for an isobaric process, the heat \(Q\) is given by:

\[ Q = nC_P\Delta T = \frac{7}{5}\Delta U \]

Using the relation between work and internal energy change for an isobaric process:

\[ W = \frac{2}{5}Q \]

Substituting the given value of \(W\):

\[ 200 = \frac{2}{5}Q \]

Solving for \(Q\):

\[ Q = \frac{5}{2} \times 200 = 500 \, \text{J} \]

Conclusion: The heat given to the gas during the process is \(700 \, \text{J}\).