Question

For the transformation of 1 mole of an ideal gas from an initial temperature of 27$^\circ$C and an initial pressure of 1 atm to a final temperature of 327$^\circ$C and a final pressure of 17 atm, the change in enthalpy, $\Delta H$, is _ _ _ kJ

• A. 6.30
• B. 8.50
• C. 11.94
• D. 17.61

Hint:

For ideal gases, enthalpy depends only on temperature and not on pressure

Answer 1

The Correct Option is C

Step 1: Relation for enthalpy change.
For an ideal gas, enthalpy depends only on temperature
\[ \Delta H = \int_{T_1}^{T_2} C_p \, dT \]

Step 2: Substitute given heat capacity expression.
\[ C_p = a + bT \]
Thus,
\[ \Delta H = \int_{T_1}^{T_2} (a + bT)\, dT \]

Step 3: Perform integration.
\[ \Delta H = a(T_2 - T_1) + \frac{b}{2}(T_2^2 - T_1^2) \]

Step 4: Convert temperatures to Kelvin.
\[ T_1 = 27 + 273 = 300 \, K, \quad T_2 = 327 + 273 = 600 \, K \]

Step 5: Substitute values.
\[ \Delta H = 20.9(600 - 300) + \frac{0.042}{2}(600^2 - 300^2) \]
\[ = 20.9 \times 300 + 0.021 (360000 - 90000) \]
\[ = 6270 + 0.021 \times 270000 \]
\[ = 6270 + 5670 = 11940 \, J \]

Step 6: Convert to kJ.
\[ \Delta H = 11.94 \, \text{kJ} \]

Step 7: Conclusion.
\[ \boxed{11.94 \, \text{kJ}} \]