105 calories of heat is required to raise the temperature of 3 moles of an ideaol gas at constant pressure from` 30^(@)C` to `35^(@)C.` The amount of heat required in calories to raise the temperature of the gas though the range `(60 ^(@)C` to `65^(@)C)` at constant volume is `(gamma = (C_(p))/( C_(v)) = 1.4)`

Options

A. 50 cal
B. 75 cal
C. 70 cal
D. 90 cal

Answer

To solve the problem, we need to determine the amount of heat required to raise the temperature of an ideal gas from 60°C to 65°C at constant volume. We will use the information provided about the heat required at constant pressure to find the necessary values.

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- Heat required at constant pressure (Q) = 105 calories
- Number of moles (n) = 3 moles
- Initial temperature (T1) = 30°C
- Final temperature (T2) = 35°C
- Change in temperature (ΔT) = T2 - T1 = 35°C - 30°C = 5°C
- Given ratio: \( \gamma = \frac{C_p}{C_v} = 1.4 \)

2. **Finding \( C_p \)**:
- The formula for heat at constant pressure is:
\[ Q = n C_p \Delta T \]
- Substituting the known values:
\[ 105 = 3 C_p \times 5 \]
- Simplifying:
\[ 105 = 15 C_p \]
- Solving for \( C_p \):
\[ C_p = \frac{105}{15} = 7 \text{ cal/mol°C} \]

3. **Finding \( C_v \)**:
- Using the relationship \( \gamma = \frac{C_p}{C_v} \):
\[ 1.4 = \frac{7}{C_v} \]
- Rearranging to find \( C_v \):
\[ C_v = \frac{7}{1.4} = 5 \text{ cal/mol°C} \]

4. **Calculating Heat at Constant Volume**:
- Now we need to find the heat required to raise the temperature from 60°C to 65°C at constant volume.
- The change in temperature (ΔT) for this process:
\[ \Delta T = 65°C - 60°C = 5°C \]
- The formula for heat at constant volume is:
\[ Q = n C_v \Delta T \]
- Substituting the values:
\[ Q = 3 \times 5 \times 5 \]
- Calculating:
\[ Q = 3 \times 5 = 15 \]
\[ Q = 15 \times 5 = 75 \text{ calories} \]

5. **Final Answer**:
- The amount of heat required to raise the temperature of the gas from 60°C to 65°C at constant volume is **75 calories**.