Question

If the pressure of an ideal gas contained in a closed vessel is increased by 0.5%, the increase in temperature is $ 2{}^\circ C $ . The initial temperature of the gas is:

• A. $ 27{}^\circ C $
• B. $ 127{}^\circ C $
• C. $ 300{}^\circ C $
• D. $ 400{}^\circ C $

Answer 1

Answer: (B)

Applying Gay Lussacs gas law at constant volume $ P\propto T $ Hence, $ \frac{{{P}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}}{{{T}_{2}}} $ ?(i) Suppose, initial pressure $ {{P}_{1}}=P $ Final pressure $ {{P}_{2}}=P+\frac{0.5P}{100} $ $ =1.005\,P $ and $ {{T}_{1}}=T{{\,}^{o}}C $ and $ {{T}_{2}}=(T+2){{\,}^{o}}C $ So, putting the values in Eq. (i), we get $ \frac{P}{T}=\frac{1.005P}{T+2} $ $ \Rightarrow $ $ T+2=1.005\,T $ $ \Rightarrow $ $ 1.005\,T=2 $ $ \Rightarrow $ $ T=\frac{2}{1.005}=400K $ $ =400-273=127{{\,}^{o}}C $