A) 200 K
B) 100 K
C) \[200{}^\circ C\]
D) \[100{}^\circ C\]
Correct Answer: B
Solution :
According to Gay Lussacs or pressure law \[P\propto T\] or \[\frac{P}{T}=cons\tan t\] \[\therefore \] \[\frac{{{P}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}}{{{T}_{2}}}\] ?(i) Given, \[{{P}_{1}}=P,{{P}_{2}}=P+\frac{2}{100}P=\frac{51P}{50},{{T}_{2}}=T+2\] From Eq. (i), we have \[\frac{P}{T}=\frac{51P/50}{T+2}\] or \[T+2=\frac{51T}{50}\] or \[50T+100=51T\] \[\therefore \] \[T=100K\]
You need to login to perform this action.
You will be redirected in
3 sec
