Question

One mole of an ideal gas at 300k and 5 atm is expanded adiabatically to a final pressure of 2 atm against external pressure of 2 atm find final temperature of the gas

Answer 1

Hello dear,

● Answer -
T2 = 180 K

● Explaination -
# Given -
n = 1
P1 = 5 atm
P2 = 2 atm
Pext = 2 atm
T = 300 K

# Solution -
Initially,
V1 = nRT / P1
V1 = 1 × 0.0821 × 300 / 5
V1 = 4.926 L

Finally,
V2 = nRT / P2
V2 = 1 × 0.0821 × 300 / 2
V2 = 12.315 L

For an adiabatic reaction,
Cv = Pext (V1-V2) / (T2-T1) ...(1)

But for monoatomic gas,
Cv = R / (γ-1)
Cv = R / (5/3-1)
Cv = 3R/2

Putting this in eqn(1)
3R/2 = Pext (V1-V2) / (T2-T1)
T2-T1 = Pext (V1-V2) / 1.5R
T2-300 = 2 (4.926-12.315) / (1.5×0.0821)
T2-300 = -120
T2 = 300-120 = 180 K

Therefore, final temperature in kelvin is 180 K.

Thanks..

Answer 2

Answer:

T2=180k

Explanation:

Given

n=1

P1=5 atm

P2=2 atm

Pext=2 atm

T=300k

Initially,

V1=nRT/P1

V1=1*0.0821*300/5

V1=4.926L

Finally,

V2=nRT/P2

V2=1*0.0821*300/2

V2=12.315L

For an adiabatic reaction,

CV=Pext(V1-V2)/(T2-T1)...(1)

But for monoatomic gas,

CV=R(gamma-1)

CV=R(5/3-1)

CV=3R/2

Putting this in equation 1

3R/2=Pext(V1-V2)(T2-T1)

T2-T1=Pext(V1-V2)/1.5R

T2-300=2(4.926-12.315)/(1.5*0.0821)

T2-300=-120

T2=300-120=180 k

Therefore ,final temperature in Kelvin scale is 180 k.

Hope it helps you friend please mark as brainiliest.