A) \[0.492\]atm
B) \[49.2\]atm
C) \[4.92\]atm
D) \[0.0492\]atm
Correct Answer: A
Solution :
Number of mol. of \[He=\frac{0.4}{4}=0.1\] Number of mol. of oxygen \[=\frac{1.6}{32}=0.05\] Number of mol of nitrogen \[=\frac{1.4}{28}=0.05\] Total mol. in the \[10.0\text{ }L\]cylinder at \[{{27}^{o}}C\] \[=0.1+0.05+0.05=0.2\text{ }mol\] \[P=\frac{nRT}{V}\] \[=\frac{0.2\times 0.082\times 300}{10}=0.492\,atm\]
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