Question

A 2.00 L sample of gas is cooled from 91*C to a temperature at which its volume is 0.60 L. What is this new temperature? Assume no change in pressure of the gas. in *C

Answer 1

We know that ideal gas equation is PV = nRT

Where

P = pressure , V = volume , n = number of moles , R = gas constant , T = temperature

As the pressure is kept constant and the gas remains the same , so V is proportional to T

V / V' = T / T'

Where

V =- initial volume = 2.00 L

V ' = final volume = 0.60 L

T = initial temperature = 91 oC = 91+273 = 364 K

T' = final temperature = ?

Plug the values we get ,

T' = (T x V') / V

   = (364 x 0.60)/ 2.00

    = 109.2K

     = 109.2 - 273 ^oC

     = -163.8 ^oC

Therefore the final temperature is -163.8 ^oC