Question

2
58. A vessel is filled with an ideal gas at a pressure
of 20 atm and is at a temperature of 27°C.
One half of the mass of the gas is removed
from the vessel and the temperature of the
remaining gas is increased to 87°C. At this
temperature the pressure of the gas will be​

Answer 1

Answer:

12atm will be the pressure of gas

Answer 2

Let volume of vessel is V.

initial number of mole of gas is n.

case 1 : temperature, T = 27°C = 273 + 27 = 300K and pressure, P = 20atm

from gas equation, PV = nRT

or, V = nR(300)/20atm......(1)

case 2 : when half of the mass of gas is removed. then number of mole of gas will be halved. e.g., nf = n/2,

temperature, Tf = 87° C = 273 + 87 = 360K, let pressure is P'.

it is happening in same vessel so, volume of gas remains constant i.e., V

now, V = nfRTf/P'

or, V = (n/2)R(360)/P' .....(2)

from equations (1) and (2),

nR(300)/20 = (n/2)R(360)/P'

or, 15 = 180/P'

or, P' = 12atm

hence, the pressure of gas will be 12atm