Question
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 for a gas is 20 J/mole - K then calculate ΔH (kJ/mol) for the process?
R = 8J/mol - K
A. 6
B. 10
C. 2.8
D. 2
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
Where H is the enthalpy, U is the internal energy, P is the pressure and V is the volume
ΔU=nCvΔT
U is the internal energy, n is the moles, Cv is the heat capacity at constant volume and T is the temperature.
Complete step by step answer:
From the question we see that, ΔT
Pressure P1 = 2 bar, volume V1 = 5 litre.
Pressure P2 = 10 bar, volume V2 = 5 litre
So, the volume remains constant in the process.
Now, from the above formula, we have,
ΔH=ΔU+PΔV+VΔP
As the volume remains constant, the change in volume is equal is zero, therefore,
ΔV=0 And PΔV=0,
so, we have, ΔH=ΔU+VΔP.
The change in internal energy of the gas is a function of the temperature and is mathematically expressed as,
ΔU=nCvΔT
Where, n is the number of moles of the gas,Cv is the specific heat capacity at constant volume and ΔTis the change in temperature.
Therefore from the above equation, by putting the values of Cv and we get, considering one mole of gas,
ΔU=1×20×(400−300)
ΔU=2000J/mol
Putting the value of ΔU, V and ΔP
ΔH=ΔU+VΔP
⇒ΔH=2000+[5×(1000−200)]
Solving this, we get:
⇒ΔH=2000+4000
⇒ΔH=6000J/mol = 6 kJ/mol
Hence the correct answer is option A.
Note:
For ideal gases, Cv - Cp = R, where Cv is the specific heat capacity of an gas at constant volume and Cp is the specific heat capacity at constant pressure.
The enthalpy change of a reaction can be written as:
ΔH=nCpΔT