An ideal diatomic gas undergoes a thermodynamic process as shown in indicator p-v diagram, heat input during process a to b is
A.12pv
B.45pv
C.57pv
D.47pv
Answers
Answer:
The answer will be 5nRT0/2
Explanation:
According to the problem the thermodynamic process of diatomic gas is given.
Now according to the figure we can say that the Process from a to b is isochoric,
Therefore Δv = 0 and Δ w = 0
As the first law of thermodynamics says,
ΔQ = Δu +Δw => ΔQ =Δu
Now we know that Δu = nCvΔT = nR/γ-1ΔT= n R/(7/5-1)ΔT
Therefore ΔQ = 5nRΔT/2
Now let assume that at point a the elements will be(P0, V0,T0)
The equation for gas,
PV = nRT
Now the pressure is being double at a-b process , volume is constant therefore the temperature will be doubled.
Therefore, at b point the temperature will be 2T0
Therefore,
ΔQ = 5nR(2T0- T0)/2
Therefore ΔQ = 5nRT0/2
Hence we can say that the heat input during process a to b is 5nRT0/2
Explanation:
Explanation:
Gas. CP. CV. $=CP/CV
Mono. 5/2R 3/2R 5/3
Di. 7/2R 5/2R 7/5
Tri. 4R 3/2R. 4/3
