When gas is given heat `DeltaQ`, a part of heat energy is utilized into work done W by gas and the remaining part is utilized to change in internal energy. An ideal diatomic gas is heated at constant pressure, the ratio of the internal energy change to heat energy supplied, is

Options

A. `(3)/(7)`
B. `(2)/(5)`
C. `(5)/(7)`
D. `(3)/(5)`

Answer

To solve the problem, we need to find the ratio of the change in internal energy (\( \Delta U \)) to the heat energy supplied (\( \Delta Q \)) for an ideal diatomic gas heated at constant pressure.

### Step-by-Step Solution:

1. **Understanding the First Law of Thermodynamics**:
The first law of thermodynamics states that the heat added to the system (\( \Delta Q \)) is equal to the change in internal energy (\( \Delta U \)) plus the work done by the system (\( W \)):
\[ \Delta Q = \Delta U + W \]

2. **Identifying Work Done at Constant Pressure**:
For an ideal gas at constant pressure, the work done (\( W \)) when the volume changes from \( V_1 \) to \( V_2 \) is given by:
\[ W = P \Delta V = P (V_2 - V_1) \]
Using the ideal gas law, we can express this in terms of temperature changes.

3. **Using the Ideal Gas Law**:
The ideal gas law states:
\[ PV = nRT \]
At constant pressure, we can relate the initial and final states:
\[ P V_1 = n R T_1 \quad \text{and} \quad P V_2 = n R T_2 \]
Thus, we can write:
\[ W = nR(T_2 - T_1) = nR \Delta T \]

4. **Calculating Change in Internal Energy**:
For a diatomic gas, the change in internal energy is given by:
\[ \Delta U = \frac{f}{2} n R \Delta T \]
where \( f \) is the degrees of freedom. For a diatomic gas, \( f = 5 \):
\[ \Delta U = \frac{5}{2} n R \Delta T \]

5. **Substituting into the First Law**:
Now substituting \( W \) and \( \Delta U \) into the first law equation:
\[ \Delta Q = \Delta U + W \]
becomes:
\[ \Delta Q = \frac{5}{2} n R \Delta T + n R \Delta T \]
Simplifying this gives:
\[ \Delta Q = \left(\frac{5}{2} + 1\right) n R \Delta T = \frac{7}{2} n R \Delta T \]

6. **Finding the Ratio \( \frac{\Delta U}{\Delta Q} \)**:
Now we can find the ratio of the change in internal energy to the heat energy supplied:
\[ \frac{\Delta U}{\Delta Q} = \frac{\frac{5}{2} n R \Delta T}{\frac{7}{2} n R \Delta T} \]
The \( n R \Delta T \) terms cancel out:
\[ \frac{\Delta U}{\Delta Q} = \frac{5}{7} \]

### Final Answer:
The ratio of the change in internal energy to the heat energy supplied is:
\[ \frac{\Delta U}{\Delta Q} = \frac{5}{7} \]