A thermally insulated vessel containing a gas whose molar mass is equal to M and the ratio of specific heats cP /cV = γ moves with a velocity v. Find the change in the temperature T of the gas resulting from the sudden stoppage of the vessel.

∆T = Mv^2Y/2Cp

Explanation:

Formula for Kinetic energy of the vessel = 1/2mv^2

Increase in internal energy Δu = nCVΔT

where n is the number of moles of the gas in vessel.

When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas

We say

1/2mv^2 = ∆u

1/2mv^2 = nCv∆T

Since n = m/M

1/2mv^2 = mCv∆T/M

Making ∆T subject of the formula we have

∆T = Mv^2/2Cv

Multiple the RHS by Cp/Cp

∆T = Mv^2/2Cv *Cp/Cp

Since Y = Cp/CV

∆T = Mv^2Y/2Cp k

Since CV = R/Y - 1

We could also have

∆T = Mv^2(Y - 1)/2R k

ΔT = (Y-1/2R) Mv² Kelvin, whereΔT = Change in temperature

Explanation:

Kinetic energy of vessel = 1/2mv², Change in internal energy(ΔU) = nCvΔT

n = number of moles of gas in vessel

When the vessel is sudeenly stopped, its kinetic energy causes a rise in the temperature of the gas

∴ 1/2mv² = ΔU

ΔU = nCVΔT

CP -CV = R,  from  CP/CV = Y, CP = CV Y

CP - CV = R

CVY - CV = R

CV(Y-1) = R

CV = R/Y-1

Recall that 1/2mv² = ΔU = nCVΔT

nCVΔT = 1/2mv²

n( R/Y-1)ΔT = 1/2mv²               ⇔ substituting the value of CV above

from n= mass(m)/Molar mass(M), ⇔ m = nM

we have ,

n (R/Y-1)ΔT = 1/2nMv²

canceling out n

∴ ΔT = (Y-1/2R) Mv² Kelvin  , the change in temperature

b) phosphorus pentoxide