Five moles of hydrogen gas are heated from `30^(@)C` to `60^(@)C` at constant pressure. Heat given to the gas is (given `R = 2 cal//mol` degrees)

Options

A. `750 cal`
B. `630 cal`
C. `1050 cal`
D. `1470 cal`

Explanation

To solve the problem of calculating the heat given to five moles of hydrogen gas heated from \(30^\circ C\) to \(60^\circ C\) at constant pressure, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Variables**:
- Number of moles, \( n = 5 \) moles
- Initial temperature, \( T_1 = 30^\circ C = 303 \, K \) (convert to Kelvin if necessary)
- Final temperature, \( T_2 = 60^\circ C = 333 \, K \) (convert to Kelvin if necessary)
- Change in temperature, \( \Delta T = T_2 - T_1 = 60 - 30 = 30^\circ C \)
- Given \( R = 2 \, \text{cal/mol} \cdot \text{degree} \)

2. **Calculate \( C_p \) for Hydrogen**:
- Hydrogen is a diatomic gas, so we can use the formula for \( C_p \):
\[ C_p = \frac{\gamma}{\gamma - 1} R \]
- For diatomic gases, \( \gamma = \frac{7}{5} \).
- Substitute \( \gamma \) and \( R \):
\[ C_p = \frac{7/5}{(7/5) - 1} \cdot 2 \]
- Simplify the denominator:
\[ (7/5) - 1 = (7/5) - (5/5) = \frac{2}{5} \]
- Now calculate \( C_p \):
\[ C_p = \frac{7/5}{2/5} \cdot 2 = \frac{7}{2} \cdot 2 = 7 \, \text{cal/mol} \cdot \text{degree} \]

3. **Calculate Heat \( \Delta Q \)**:
- Use the formula for heat at constant pressure:
\[ \Delta Q = n C_p \Delta T \]
- Substitute the values:
\[ \Delta Q = 5 \cdot 7 \cdot 30 \]
- Calculate:
\[ \Delta Q = 5 \cdot 7 = 35 \]
\[ \Delta Q = 35 \cdot 30 = 1050 \, \text{calories} \]

4. **Conclusion**:
- The heat given to the gas is \( \Delta Q = 1050 \, \text{calories} \).

### Final Answer
The correct option is 3, which states that the heat given to the gas is **1050 calories**.