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One mole of an ideal gas, Cp = 4 R at 300 K Is expanded adiabatically to n times the original volume. What is value of n if temperature falls by 150°?
Correct answer is '8'. Can you explain this answer?
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Verified Answer
One mole of an ideal gas, Cp = 4 R at 300 K Is expanded adiabatically ...
Cp - Cv = R
4R - Cv = R
Cv = 3R
γ = Cp/Cv = 4/3
T1/T2 = (v2/v1)(4/3 - 1)
300/150 = (v2/v1)1/3
2 = (v2/v1)1/3
8 = v2/v1
v2 = 8v1
Therefore n = 8
4R - Cv = R
Cv = 3R
γ = Cp/Cv = 4/3
T1/T2 = (v2/v1)(4/3 - 1)
300/150 = (v2/v1)1/3
2 = (v2/v1)1/3
8 = v2/v1
v2 = 8v1
Therefore n = 8
Most Upvoted Answer
One mole of an ideal gas, Cp = 4 R at 300 K Is expanded adiabatically ...
Solution:
Given,
Cp = 4R,
T1 = 300K,
adiabatic expansion,
Temperature falls by 150K
We know that for adiabatic process,
PVγ = constant,
where γ = Cp/Cv
For one mole of ideal gas,
PV = RT
So,
P = RT/V
Substituting this in the adiabatic equation,
(RT/V) * Vγ = constant
Tγ-1 * V1-γ = constant
As the process is adiabatic,
γ = Cp/Cv
For an ideal gas,
Cp - Cv = R
So,
γ = Cp/Cp - R/Cp
γ = (Cp - R)/Cp
γ = 3/4
As the process is adiabatic,
T2 = T1 (V1/V2)γ
T2/T1 = (V1/V2)γ
(T2/T1)1/γ = V1/V2
n = V2/V1
n = (T1/T2)1/γ
n = (300/150)4/3
n = 2 * 2 * 2
n = 8
Therefore, the value of n is 8 if the temperature falls by 150K during the adiabatic expansion of one mole of an ideal gas, Cp = 4R at 300K.
Given,
Cp = 4R,
T1 = 300K,
adiabatic expansion,
Temperature falls by 150K
We know that for adiabatic process,
PVγ = constant,
where γ = Cp/Cv
For one mole of ideal gas,
PV = RT
So,
P = RT/V
Substituting this in the adiabatic equation,
(RT/V) * Vγ = constant
Tγ-1 * V1-γ = constant
As the process is adiabatic,
γ = Cp/Cv
For an ideal gas,
Cp - Cv = R
So,
γ = Cp/Cp - R/Cp
γ = (Cp - R)/Cp
γ = 3/4
As the process is adiabatic,
T2 = T1 (V1/V2)γ
T2/T1 = (V1/V2)γ
(T2/T1)1/γ = V1/V2
n = V2/V1
n = (T1/T2)1/γ
n = (300/150)4/3
n = 2 * 2 * 2
n = 8
Therefore, the value of n is 8 if the temperature falls by 150K during the adiabatic expansion of one mole of an ideal gas, Cp = 4R at 300K.
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Community Answer
One mole of an ideal gas, Cp = 4 R at 300 K Is expanded adiabatically ...
Cp - Cv = R
4R - Cv = R
Cv = 3R
Gamma = x = Cp/Cv = 4/3
T1/T2 = (v2/v1)^(4/3 - 1)
300/150 = (v2/v1)^1/3
2 = (v2/v1)^1/3
8 = v2/v1
v2 = 8v1
Therefore n = 8
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One mole of an ideal gas, Cp = 4 R at 300 K Is expanded adiabatically to n times the original volume. What is value of n if temperature falls by 150°?Correct answer is '8'. Can you explain this answer?
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