A) \[27{}^\circ C\]
B) \[127{}^\circ C\]
C) \[300{}^\circ C\]
D) \[400{}^\circ C\]
Correct Answer: B
Solution :
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 \[\frac{P}{T}=\frac{1.005P}{T+2}\] \[\Rightarrow \] \[T+2=1.005\,T\] \[\Rightarrow \] \[1.005\,T=2\] \[\Rightarrow \] \[T=\frac{2}{1.005}=400K\] \[=400-273=127{{\,}^{o}}C\]
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