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 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

Options

A. 10atm
B. 12atm
C. 14atm
D. 8atm

Answer

B

Explanation

`P_(1)/T_(1) = P_(2)/T_(2)` [`because` V=Constant]
`20/(27+273) = P_(2)/(87+273)`
`therefore P_(2)= 20/300 xx 360 = 24atm`
Since the mass of the gas is halved,
Final pressure `= P_(2)/2 = 12 atm`