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: Enthalpy of an ideal gas.
For an ideal gas the enthalpy depends only on temperature, so pressure does not enter.
\[ \Delta H = \int_{T_1}^{T_2} C_p \, dT \]

Step 2: Set the temperatures.
\[ T_1 = 300 \text{ K}, \quad T_2 = 600 \text{ K} \]

Step 3: Integrate $C_p = a + bT$ and put in values.
\[ \Delta H = a(T_2-T_1) + \frac{b}{2}(T_2^2 - T_1^2) = 6270 + 5670 = 11940 \text{ J} \]

Step 4: Answer.
\[ \boxed{11.94 \, \text{kJ}} \]