If at 60 ° C and 80 cm of mercury pressure, a definite mass of a gas is compressed slowly, then the final pressure of the gas if the final volume is half of the initial volume (γ = 3/2) is

Question

If at 60 ° C and 80 cm of mercury pressure, a definite mass 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 (A)  120 cm of Hg  (B)  140 cm of Hg  (C)  160 cm of Hg  (D)  180 cm of Hg

Tardigrade Admin · Accepted Answer

Here, P1 = 80 cm of Hg . If the gas is compressed slowly, then the process is isothermal. In this case P1V1 = P2V2 ∴ P2 =( P1V1/V2) = (80× V1/(V1)2) = 160 cm of Hg (∵ V2 =(V1/2))

Q. If at $6 0^{\circ} C$ and $80 c m$ of mercury pressure, a definite mass of a gas is compressed slowly, then the final pressure of the gas if the final volume is half of the initial volume $(γ = 3/2)$ is

$120 c m$ of $H g$

$140 c m$ of $H g$

$160 c m$ of $H g$

$180 c m$ of $H g$

Here, $P_{1} = 80 c m$ of $H g$ .
If the gas is compressed slowly, then the process is isothermal.
In this case $P_{1} V_{1} = P_{2} V_{2}$
$∴ P_{2} = \frac{P _{1} V _{1}}{V _{2}}$
$= \frac{80 \times V _{1}}{\frac{V _{1}}{2}}$
$= 160 c m$ of $H g (∵ V_{2} = \frac{V _{1}}{2})$

Q. If at 60∘C and 80cm of mercury pressure, a definite mass of a gas is compressed slowly, then the final pressure of the gas if the final volume is half of the initial volume (γ=3/2) is

120cm of Hg

140cm of Hg

160cm of Hg

180cm of Hg