Question

prove that heat capacity at constant volume by 0.0822atmdm3.k-1.mol-1​

Answer 1

Answer:

Explanation:

The initial values are n=3 mole,T  

1

​  

=200K,P  

1

​  

=2.0 atm,C  

v

​  

=27.5 JK  

−1

mol  

−1

 

The final values (after compression) are T  

2

​  

=250 K,P  

2

​  

=?

The heat capacity at constant pressure is C  

p

​  

=27.5+8.314 JK  

−1

mol  

−1

=35.814 JK  

−1

mol  

−1

 

Hence, V  

2

​  

=11.81 litre

The work done is =  

γ−1

nR

​  

[T  

2

​  

−T  

1

​  

]=  

0.3

3×8.314

​  

×[250−200]=+4157J=+4.157 kJ

Since the process is adiabatic, no heat is transferred. q=0

Internal energy decreases as work is done. Hence, ΔU=w=4.157 kJ

Also, the expression for the enthalpy change is ΔH=n×C  

p

​  

×ΔT

Substitute values in the above expression.

ΔH=3×35.814×50=5372.1 J=5.372 kJ