The amount of heat required to raise the temperature of a diatomic gas by 1ºC at constant pressure is `Q_(p)` and at constant volume is `Q_(v)`. The amount of heat which goes as internal energy of the gas in the two cases is nearly

Options

A. `Q_(P)` and `Q_(v)`
B. `0.71 Q_(P)` and `0.71 O_(v)`
C. `0.71 Q_(P)` and `Q_(V)`
D. `0.7 Q_(P)` and `0.9 Q_(V)`

Answer

To solve the problem, we need to determine the amount of heat that contributes to the internal energy of a diatomic gas when its temperature is raised by 1ºC at constant pressure and constant volume. Let's break this down step by step.

### Step 1: Understand the Heat Transfer at Constant Pressure and Constant Volume
1. **At Constant Pressure (Qp)**: The heat transfer at constant pressure is given by the formula:
\[ Q_p = n C_p \Delta T \]
where \(C_p\) is the specific heat at constant pressure, \(n\) is the number of moles, and \(\Delta T\) is the change in temperature.

2. **At Constant Volume (Qv)**: The heat transfer at constant volume is given by the formula:
\[ Q_v = n C_v \Delta T \]
where \(C_v\) is the specific heat at constant volume.

### Step 2: Determine the Specific Heats for a Diatomic Gas
For a diatomic gas:
- The degrees of freedom \(f\) is 5.
- The specific heats are calculated as follows:
\[ C_p = \frac{f}{2} R + R = \frac{5}{2} R + R = \frac{7}{2} R \]
\[ C_v = \frac{f}{2} R = \frac{5}{2} R \]

### Step 3: Calculate Qp and Qv for a Temperature Change of 1ºC
1. **Calculate Qp**:
\[ Q_p = n C_p \Delta T = n \left(\frac{7}{2} R\right) (1) = \frac{7}{2} n R \]

2. **Calculate Qv**:
\[ Q_v = n C_v \Delta T = n \left(\frac{5}{2} R\right) (1) = \frac{5}{2} n R \]

### Step 4: Relate Internal Energy Change to Qv and Qp
The change in internal energy (\(\Delta U\)) for an ideal gas is given by:
\[ \Delta U = n C_v \Delta T \]
Since we already calculated \(Q_v\) as:
\[ Q_v = \frac{5}{2} n R \]
we can say:
\[ \Delta U = Q_v \]

### Step 5: Express Internal Energy Change in Terms of Qp
To express \(\Delta U\) in terms of \(Q_p\):
1. From the earlier equations, we have:
\[ \Delta U = n C_v \Delta T = \frac{5}{2} n R \]
2. We also have:
\[ Q_p = \frac{7}{2} n R \]
3. To express \(\Delta U\) in terms of \(Q_p\):
\[ \Delta U = \frac{5}{7} Q_p \]

### Conclusion
The amount of heat that goes into the internal energy of the gas in the two cases is:
- At constant volume, \(\Delta U = Q_v\)
- At constant pressure, \(\Delta U = \frac{5}{7} Q_p\)

### Final Answer
The internal energy change in terms of \(Q_p\) is approximately \(0.71 Q_p\).

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