Question

If at $ 60 ^{\circ } 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 $

Answer 1

Answer: (C)

Here, $ P_1 = 80 \,cm $ of $ Hg $ .
If the gas is compressed slowly, then the process is isothermal.
In this case $ P_1V_1 = P_2V_2 $
$ \therefore P_{2} =\frac{ P_{1}V_{1}}{V_{2}} $
$ = \frac{80\times V_{1}}{\frac{V_{1}}{2}} $
$ = 160\, cm $ of $ Hg \left(\because V_{2} =\frac{V_{1}}{2}\right) $