Question

A bottle at 325 K contains an ideal gas at a pressure of 1.613×105 Pa . The rubber stopper closing the bottle is removed. The gas expands adiabatically against \(P_{external} \ = \\) 1.168×105 Pa , and some gas is expelled from the bottle in the process. When P = Pexternal , the stopper is quickly replaced. The gas remaining in the bottle slowly warms up to 325 K What is the final pressure in the bottle for a monatomic gas, for which CV,m = 3R/2?

Answer 1

Given,

The initial temperature T = 325 K

Pressure P = 1.613*10⁵ Pa

After the gas expands adiabatically,

P' = 1.168 * 10⁵ Pa

Temperature = T'

For a adiabatic process,

P^1- * T= constant

P^1- * T= P'^1- * T'

T / T' = (P' / P)^{(1-) /} .........(1)

For the gas inside the bottle after stopper is quickly replaced,

Pi = P'

Ti = T'

Tf = T

Here, volume is constant,

Hence, P' / T' = Pf / T

Pf = P' * (T / T')

Put the value of T / T' from eq(1)

Pf = P' * (P' / P)^{(1-) /}   

Pf = P'^1/* P^{-1 /}

For monatomic gas CV,m = 3R/2, = 5/3),

Pf = (1.168*10⁵ )^3/5 * (1.613*10⁵)^{(5/3 - 1) / 5/3}

Pf = (1.0976*10³⁾ * ((1.613*10⁵)^(2/5)

Pf = (1.0976*10³⁾ *(1.2107*10²)

Pf = 1.3288*10⁵ Pa