Question

Heat is supplied at constant pressure to a diatomic gas. The part of this heat that was utilized to increase its internal energy is:

• A. \( \frac{4}{5} \)
• B. \( \frac{5}{7} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{5}{6} \)

Hint:

Understanding the specific heat ratios and how they relate to heat partitioning in gases is crucial in thermodynamics, especially when considering molecular complexity.

Answer 1

The Correct Option is B

For a diatomic gas, the heat supplied at constant pressure \( Q \) is related to the change in temperature \( \Delta T \) by:

\[ Q = n C_P \Delta T \]

where:

• \( n \) is the number of moles of the gas,
• \( C_P \) is the specific heat at constant pressure.

The total change in internal energy \( \Delta U \) for a diatomic gas is given by:

\[ \Delta U = n C_V \Delta T \]

where \( C_V \) is the specific heat at constant volume.

The relationship between \( C_P \) and \( C_V \) for an ideal gas is:

\[ C_P = C_V + R \]

For a diatomic ideal gas, the ratio \( \gamma = \frac{C_P}{C_V} \) is given by:

\[ \gamma = \frac{7}{5} \]

Thus, we have:

\[ C_P = \frac{7}{5} C_V \]

Now, the part of the heat that increases the internal energy is the heat utilized to increase the temperature at constant volume, which is the change in internal energy. The fraction of the heat supplied that increases internal energy is given by:

\[ \frac{\Delta U}{Q} = \frac{n C_V \Delta T}{n C_P \Delta T} = \frac{C_V}{C_P} = \frac{5}{7} \]

Thus, the fraction of heat utilized to increase the internal energy is \( \frac{5}{7} \).

Therefore, the correct answer is Option (2), \( \frac{5}{7} \).

Answer 2

To solve the problem, we need to determine the portion of heat supplied to a diatomic gas at constant pressure that increases its internal energy.

1. Understanding the Problem:
For an ideal gas undergoing a process at constant pressure, the heat supplied is partially used to increase the internal energy of the gas, and the remaining portion does work in expanding the gas. The relationship between heat (\( Q \)), work done (\( W \)), and the change in internal energy (\( \Delta U \)) is given by the first law of thermodynamics:

\[ Q = \Delta U + W \] At constant pressure, the work done by the gas is given by: \[ W = P \Delta V \] For a diatomic gas, the change in internal energy \( \Delta U \) is related to the temperature change \( \Delta T \) by: \[ \Delta U = \frac{5}{2} n R \Delta T \] where \( n \) is the number of moles and \( R \) is the universal gas constant. The heat supplied \( Q \) at constant pressure is given by: \[ Q = n C_P \Delta T \] where \( C_P \) is the heat capacity at constant pressure, and for a diatomic gas, \( C_P = \frac{7}{2} R \). The fraction of heat used to increase the internal energy is the ratio: \[ \frac{\Delta U}{Q} = \frac{\frac{5}{2} n R \Delta T}{n \frac{7}{2} R \Delta T} = \frac{5}{7} \]

2. Conclusion:
The fraction of heat used to increase the internal energy is:

Final Answer:
The correct option is (B) \( \frac{5}{7} \).