The temperature of 2 moles of an ideal gas is raised from `27^(@)C`to `77^(@)C`. What is the value for`DeltaH - DeltaU` for the process ? `( R= 8.3 J K^(-1)mol^(-1))`

Options

A. 415 J
B. 830 J
C. 1660 J
D. None of these

Explanation

To solve the problem, we need to find the value of \( \Delta H - \Delta U \) for the given process involving an ideal gas.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- The initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- The final temperature \( T_2 = 77^\circ C = 77 + 273 = 350 \, K \)

2. **Calculate the Change in Temperature (\( \Delta T \)):**
\[ \Delta T = T_2 - T_1 = 350 \, K - 300 \, K = 50 \, K \]

3. **Use the Relationship Between Enthalpy and Internal Energy:**
- The relationship is given by:
\[ \Delta H = \Delta U + \Delta (PV) \]
- For an ideal gas, \( PV = nRT \), so:
\[ \Delta (PV) = nR \Delta T \]

4. **Substituting into the Equation:**
- Therefore, we can express \( \Delta H - \Delta U \) as:
\[ \Delta H - \Delta U = \Delta (PV) = nR \Delta T \]

5. **Plugging in the Values:**
- Given:
- \( n = 2 \, \text{moles} \)
- \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( \Delta T = 50 \, K \)
- Now substituting the values:
\[ \Delta H - \Delta U = 2 \, \text{moles} \times 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \times 50 \, K \]

6. **Calculating the Result:**
\[ \Delta H - \Delta U = 2 \times 8.314 \times 50 = 831.4 \, \text{J} \]

### Final Answer:
\[ \Delta H - \Delta U = 831.4 \, \text{J} \]