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A certain mass of a gas at 273 k temperature and one atmospheric pressure is expanded to 3 times its original volume under adia...Asked by gourav
A certain mass of a gas at 273k temperature and one atmospheric pressure is expanded to 3 times its original volume under adiabatic conditions.Calculate the resulting temperature and pressure( take the value of (gamma)= 1.4).
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Answered by
Jai
Under adiabatic conditions (dQ = 0),
P2 / P1 = (V1 / V2)^γ
where
P1 and P2 = initial and final pressures respectively
V1 and V2 = initial and final volumes respectively
γ = heat capacity ratio = Cp/Cv
Substituting,
P2 / 1 = (V1 / 3*V1)^1.4
P2 = 0.2148 atm
Also, under adiabatic conditions,
T2 / T1 = (V1 / V2)^(γ-1)
where
T1 and T2 = initial and final absolute temperatures respectively
Substituting,
T2 / 273 = (V1 / 3*V1)^(1.4-1)
T2 = 175.92 K
hope this helps? `u`
P2 / P1 = (V1 / V2)^γ
where
P1 and P2 = initial and final pressures respectively
V1 and V2 = initial and final volumes respectively
γ = heat capacity ratio = Cp/Cv
Substituting,
P2 / 1 = (V1 / 3*V1)^1.4
P2 = 0.2148 atm
Also, under adiabatic conditions,
T2 / T1 = (V1 / V2)^(γ-1)
where
T1 and T2 = initial and final absolute temperatures respectively
Substituting,
T2 / 273 = (V1 / 3*V1)^(1.4-1)
T2 = 175.92 K
hope this helps? `u`
Answered by
gourav
Thanks Mr jai for this suggestion
Answered by
Jai
You're welcome~!
I'm a girl though, but it's alright~ :D
I'm a girl though, but it's alright~ :D
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