Question

A monoatomic gas of \( n \)-moles is heated from temperature \( T_1 \) to \( T_2 \) under two different conditions:

• A. More when heated at constant volume
• B. More when heated at constant pressure
• C. Same in both the cases
• D. Zero in both the cases

Hint:

The internal energy of an ideal gas depends only on temperature. It remains the same for a given temperature change, regardless of whether the process is at constant volume or constant pressure.

Answer 1

The Correct Option is C

Step 1: Understanding Internal Energy Change 
For an ideal gas, the internal energy \( U \) depends only on temperature and is given by: \[ \Delta U = n C_V \Delta T. \] Since internal energy is a state function, the change in internal energy depends only on the initial and final temperatures, regardless of the process. 
Step 2: Applying to Both Conditions 
- When heated at constant volume: \[ \Delta U = n C_V (T_2 - T_1). \] - When heated at constant pressure: \[ \Delta U = n C_V (T_2 - T_1). \] Since \( \Delta U \) depends only on \( T_2 - T_1 \), it is the same in both cases. 
Step 3: Conclusion 
Thus, the change in internal energy remains the same whether heating occurs at constant volume or constant pressure.

Answer 2

To determine the heat required for heating a monoatomic gas from temperature \( T_1 \) to \( T_2 \) under different conditions, let's consider the following analysis:

1. Heating at Constant Volume:

The heat \( Q_v \) required at constant volume is given by:

\( Q_v = nC_v(T_2-T_1) \)

For a monoatomic ideal gas, the molar heat capacity at constant volume \( C_v = \frac{3}{2} R \), where \( R \) is the universal gas constant.

Thus,

\( Q_v = n \left(\frac{3}{2}R\right)(T_2-T_1) = \frac{3}{2}nR(T_2-T_1) \)

2. Heating at Constant Pressure:

The heat \( Q_p \) required at constant pressure is given by:

\( Q_p = nC_p(T_2-T_1) \)

For a monoatomic ideal gas, the molar heat capacity at constant pressure \( C_p = \frac{5}{2} R \).

Thus,

\( Q_p = n \left(\frac{5}{2}R\right)(T_2-T_1) = \frac{5}{2}nR(T_2-T_1) \)

Comparison:

To compare the two, note:

- The internal energy change \( \Delta U \) for both processes depends only on the temperature difference and is given by:

\( \Delta U = nC_v(T_2-T_1) = \frac{3}{2}nR(T_2-T_1) \)

- For an ideal gas, the change in internal energy is the same for any process since it's a function of the initial and final temperatures only.

Since \( C_p - C_v = R \) and the work done at constant pressure is \( P \Delta V = nR(T_2-T_1) \), using \( Q_p \) includes this work term.

The heat added \( Q \) does differ for \( Q_p \) and \( Q_v \), but the crucial point is that the energy required (i.e., the energy change) for changing the state of a specific quantity of gas between two temperatures is fundamentally the same in terms of raising the internal energy.

Hence, the correct conclusion is that the heat required is same in both the cases to change the temperature, as the internal energy change remains constant.