Chemistry - Thermodynamics - I Question with Solution | TestHub

The enthalpy change, $\Delta H$, for a real gas undergoing a change from (3 lit, 5 atm) to (5 lit, 7 atm) is calculated using partial derivatives.

Given:

$(\frac{\partial H}{\partial P}{)}_V=5$ lit

$(\frac{\partial H}{\partial V}{)}_P=2V$ atm

The total differential for enthalpy $H$ can be written as:

$$dH={\left(\frac{\partial H}{\partial P}\right)}_{V}dP+{\left(\frac{\partial H}{\partial V}\right)}_{P}dV$$

Substituting the given values:

$$dH=5dP+2VdV$$

To find the total change in enthalpy, $\Delta H$, we integrate this expression from the initial state to the final state.

Initial state: $P_1=5$ atm, $V_1=3$ lit

Final state: $P_2=7$ atm, $V_2=5$ lit

$$\Delta H={\int }_{P_1}^{P_2}5dP+{\int }_{V_1}^{V_2}2VdV$$

$$\Delta H=5[P{]}_{P_1}^{P_2}+2{\left[\frac{V^2}{2}\right]}_{V_1}^{V_2}$$

$$\Delta H=5(P_2-P_1)+(V_2^2-V_1^2)$$

Substitute the values:

$$\Delta H=5(7-5)+(5^2-3^2)$$

$$\Delta H=5(2)+(25-9)$$

$$\Delta H=10+16$$

$$\Delta H=26atm-litre$$

The final answer is $\boxed{26atm-litre}$.