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can i get bit mesra with 110 in jee main and 89% marks in cbse board.i belong to general category.

can i get bit mesra with 110 in jee main and 89% marks in cbse board.i belong to general category.

Grade:12th pass

1 Answers

Rinkoo Gupta
askIITians Faculty 81 Points
9 years ago
Start with the definition of heat capacity at constant pressure
Cp = (?H/?T)p ((..)p indicates derivation at constant pressure)
Substitute the enthalpy by its definition
H = E + pV
Therefore:
Cp = (?E/?T)p + (?(pV)/?T)p = (?E/?T)p + p×(?V/?T)p
Because
dE = (?E/?V)t dV + (?E/?T)v dT = (?E/?V)t dV + Cv dT
The partial derivative of E with respect to temperature at constant pressure is:
(?E/?T)p = (?E/?V)t (?V/?T)p + Cv

Join both expression together:
Cp - Cv = (?E/?V)t (?V/?T)p + p×(?V/?T)p = [(?E/?V)t + p]×(?V/?T)p
q.e.d.

For further calculations the is a useful relation to evaluate the partial derivative of the energy from an equation of state.
As shown in my answer to your last question (see link):
(?E/?V)t = T×(?p/?T)v - p

For an ideal gas
pV = RT <=> p = (RT)/V <=> V = (RT)/p
(?E/?V)t = T×R/V - p = 0
(?V/?T)p = R/p
Cp - Cv = [p + 0] × R/p = R

For a Van der Waals gas
(p + a/V²) = (RT)/(V-b)
As shown before (see link):
(?E/?V)t = a/V²
The partial derivative of the volume you get by implicit differentiation
(p + a/V²) - (RT)/(V-b) = 0
-R/(V-b) +[-2a/V³ + (RT)/(V-b)²] (?V/?T)p = 0
=>
(?V/?T)p = [R/(V-b)] / [(RT)/(V-b)² -2a/V³]

Cp - Cv = [p+a/V²] × [R/(V-b)] / [(RT)/(V-b)² -2a/V³]
= [(RT)/(V-b)] × [R/(V-b)] / [(RT)/(V-b)² -2a/V³]
= R / (1 - [2a×(V-b)²]/[RTV³])
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty

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