A) 0.492 atm
B) 49.2 atm
C) 4.92 atm
D) 0.0492 atm
Correct Answer: A
Solution :
Number of moles of helium\[=\frac{0.4}{4}=0.1\] Number of moles of oxygen\[=\frac{1.6}{3.2}=0.05\] Number of moles of nitrogen \[=\frac{1.4}{28}=0.05\] Total moles in the 10.0 L cylinder at\[27{}^\circ C\] \[=(0.1+0.05+0.05)=0.2\,\,mol\] \[{{p}_{T}}=\frac{nRT}{V}=\frac{0.2\times 0.082\times 300}{10}\] \[=0.492\,atm\]
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