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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...

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 Cv{{\text{C}}_{\text{v}}} for a gas is 20 J/mole - K{\text{J/mole - K}} then calculate ΔH{{\Delta H}} (kJ/mol)\left( {{\text{kJ/mol}}} \right) for the process?
R = 8J/mol - K{\text{R = 8J/mol - K}}

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

Explanation

Solution

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:
ΔH=ΔU+PΔV+VΔP{{\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
ΔU=nCvΔT{{\Delta U = n}}{{\text{C}}_{\text{v}}}{{\Delta T}}
U is the internal energy, n is the moles, Cv{{\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, ΔT{{\Delta T}}
Pressure P1{{\text{P}}_{\text{1}}} = 2 bar, volume V1{{\text{V}}_{\text{1}}} = 5 litre.
Pressure P2{{\text{P}}_{\text{2}}} = 10 bar, volume V2{{\text{V}}_{\text{2}}} = 5 litre
So, the volume remains constant in the process.
Now, from the above formula, we have,
ΔH=ΔU+PΔV+VΔP{{\Delta H = \Delta U + P\Delta V + V\Delta P}}
As the volume remains constant, the change in volume is equal is zero, therefore,
ΔV=0{{\Delta V = 0}} And PΔV=0{{P\Delta V = 0}},
so, we have, ΔH=ΔU+VΔP{{\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,
ΔU=nCvΔT{{\Delta U = n}}{{\text{C}}_{\text{v}}}{{\Delta T}}
Where, n is the number of moles of the gas,Cv{{\text{C}}_{\text{v}}} is the specific heat capacity at constant volume and ΔT{{\Delta T}}is the change in temperature.
Therefore from the above equation, by putting the values of Cv{{\text{C}}_{\text{v}}} and we get, considering one mole of gas,
ΔU=1×20×(400300){{\Delta U}} = 1 \times 20 \times \left( {400 - 300} \right)
ΔU=2000J/mol{{\Delta U}} = 2000{\text{J/mol}}
Putting the value of ΔU{{\Delta U}}, V and ΔP{{\Delta P}}
ΔH=ΔU+VΔP{{\Delta H = \Delta U + V\Delta P}}
ΔH=2000+[5×(1000200)]\Rightarrow {{\Delta H}} = 2000 + \left[ {5 \times \left( {1000 - 200} \right)} \right]
Solving this, we get:
ΔH=2000+4000\Rightarrow {{\Delta H}} = 2000 + 4000
ΔH=6000J/mol = 6 kJ/mol\Rightarrow {{\Delta H}} = 6000{\text{J/mol = 6 kJ/mol}}

Hence the correct answer is option A.

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