If at `60^(@)` and 80 cm of mercury pressure a definite masss of a gas is compressed slowly, then the final pressure of the gas if the final volume is half of the initial volume `(gamma =(3)/(2))` is

Options

A. 120 cm of Hg
B. 140 cm of Hg
C. 160 cm of Hg
D. 180 cm of Hg

Answer

C

Explanation

Here `P_(1)=80 cm of Hg`
If the gas is compressed slowily then the fprocess is isothermal .
In this case `P_(1)V_(1)=P_(2)V_(2)`
`P_(2)=(P_(1)V_(1))/(V_(2))=(80xxV_(1))/(V_(1)//2)=160` cm of Hg