Adiabatic Slope ?

mkbh_10 · Apr 7, 2008

1. The problem statement, all variables and given/known data

The slope of an adiabatic curve will be equal to r times the slope of an isothermal curve only when we use the ideal gas equation

2. Relevant equations

3. The attempt at a solution

Adibatic process, P(V)^g = constant
Differentiating w.r.t V

(V)^gdp/dv+[g(v)^g-1]P=0

dp/dv= -[g(V)^g-1]P/v^g

dp/dv= -g P/V

Slope of an isothermal curve using P=nRT/V is -nRT/V^2 = -P/V

Therefore slope Adiabatic = g times slope isothermal

Am i Correct ??

Can we use vander waal eqn also ?

genneth · Apr 7, 2008

You've proven what you put it --- the formula pV^g=constant for adiabatic expansion uses the ideal gas law for its derivation. You have to start more generally.

mkbh_10 · Apr 7, 2008

how do i do that ?

genneth · Apr 8, 2008

Start with a general equation of state, i.e. p is some unknown function of V and T. Relate the gradients of isothermal and adiabatic curves to its values or derivatives. Set them to be proportional to each other, with a constant -g, and then solve the resulting differential equations to show that you end up with p=GVT where G is some (unknown) constant.

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Adiabatic Slope ?

mkbh_10

genneth

mkbh_10

genneth

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