A) 3 g
B) 2 g
C) 4 g
D) 5 g
Correct Answer: B
Solution :
From ideal gas equation \[pV=n\,RT\] where \[n=\frac{m}{M}\], m is weight of gas, M the molecular weight, R the gas constant. \[\therefore \] \[pV=\frac{m}{M}RT\] ... (i) and \[pV=\frac{m}{M}RT\] ... (ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{p}{p}=\frac{m}{m}\,.\,\frac{T}{T}\] \[\Rightarrow \] \[m=\frac{p/2}{p}\times \frac{400}{300}\times 6=4\,\,g\] \[\therefore \] mass of the gas that leaked out \[=6\,g-4g=2g\]You need to login to perform this action.
You will be redirected in
3 sec