Question

A non-ideal gas take is taken from state 1 (2 bar, 5 litre, 300K) to state 2 (10 bar, 5 litre, 400 K). If ${{\text{C}}_{\text{v}}}$ for a gas is 20 \[{\text{J/mole - K}}\] then calculate ${{\Delta H}}$ \[\left( {{\text{kJ/mol}}} \right)\] for the process?
\[{\text{R = 8J/mol - K}}\]

A. 6
B. 10
C. 2.8
D. 2

Answer 1

Hint: This sum can be solved from the knowledge of first law of thermodynamics, which can be stated as, the energy supplied to a system from outside in the form heat is spent in increasing the internal energy of the system and to perform external pressure-volume work.

Formula used:
 ${{\Delta H = \Delta U + P\Delta V + V\Delta P}}$
Where H is the enthalpy, U is the internal energy, P is the pressure and V is the volume
${{\Delta U = n}}{{\text{C}}_{\text{v}}}{{\Delta T}}$
U is the internal energy, n is the moles, ${{\text{C}}_{\text{v}}}$ is the heat capacity at constant volume and T is the temperature.

Complete step by step answer:
From the question we see that, ${{\Delta T}}$
Pressure \[{{\text{P}}_{\text{1}}}\] = 2 bar, volume ${{\text{V}}_{\text{1}}}$ = 5 litre.
Pressure \[{{\text{P}}_{\text{2}}}\] = 10 bar, volume ${{\text{V}}_{\text{2}}}$ = 5 litre
So, the volume remains constant in the process.
Now, from the above formula, we have,
${{\Delta H = \Delta U + P\Delta V + V\Delta P}}$
As the volume remains constant, the change in volume is equal is zero, therefore,
${{\Delta V = 0}}$ And ${{P\Delta V = 0}}$,
so, we have, ${{\Delta H = \Delta U + V\Delta P}}$.
The change in internal energy of the gas is a function of the temperature and is mathematically expressed as,
${{\Delta U = n}}{{\text{C}}_{\text{v}}}{{\Delta T}}$
Where, n is the number of moles of the gas,${{\text{C}}_{\text{v}}}$ is the specific heat capacity at constant volume and ${{\Delta T}}$is the change in temperature.
Therefore from the above equation, by putting the values of ${{\text{C}}_{\text{v}}}$ and we get, considering one mole of gas,
${{\Delta U}} = 1 \times 20 \times \left( {400 - 300} \right)$
${{\Delta U}} = 2000{\text{J/mol}}$
Putting the value of ${{\Delta U}}$, V and ${{\Delta P}}$
${{\Delta H = \Delta U + V\Delta P}}$
$ \Rightarrow {{\Delta H}} = 2000 + \left[ {5 \times \left( {1000 - 200} \right)} \right]$
Solving this, we get:
$ \Rightarrow {{\Delta H}} = 2000 + 4000$
$ \Rightarrow {{\Delta H}} = 6000{\text{J/mol = 6 kJ/mol}}$

Hence the correct answer is option A.

Note:
For ideal gases, ${{\text{C}}_{\text{v}}}{\text{ - }}{{\text{C}}_{\text{p}}}{\text{ = R}}$, where ${{\text{C}}_{\text{v}}}$ is the specific heat capacity of an gas at constant volume and ${{\text{C}}_{\text{p}}}$ is the specific heat capacity at constant pressure.
The enthalpy change of a reaction can be written as:
${{\Delta H = n}}{{\text{C}}_{\text{p}}}{{\Delta T}}$