A) 2 g
B) 3 g
C) 4 g
D) 1 g
Correct Answer: D
Solution :
From ideal gas equation, \[pV=\frac{m}{M}RT\] In\[I\]st condition, \[pV=\frac{6}{M}R(500)\] ...(i) Let\[x\]gram of hydrogen leaks out, So, \[\frac{P}{2}V=\frac{(6-x)}{M}R(300)\] ...(ii) Dividing Eqs. (ii) by Eqs. (i), we have \[\frac{1}{2}=\frac{6-x}{6}.\frac{300}{500}\] \[\frac{1}{2}=\frac{6-x}{10}\] \[\Rightarrow \] \[6-x=5\] \[\therefore \] \[x=1g\]You need to login to perform this action.
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