The pressure exerted by a sample of a gas confined in \(5.86\) \(dm ^{3}\) container at \(20^{\circ} C\) is \(4.1 atm\). What is the number of moles of the gas in the sample? \(\left( R =0.082 dm ^{3} atm \textrm {mol } ^ { - } \right.\) \(\left.1 K ^{-1}\right)\)
A. \(1.00\) B. \(2.00\) C. \(3.00\) D. \(4.00\)
Correct Answer: A
Explanation
\(V=5.86 dm ^{3}\) \(T=20^{\circ} C =20+273=293 K\) \(P=4.1 atm , R=0.082 dm ^{3} atmmol ^{-1} k ^{-1}\) \(n=\) ? from \(P V=n R T\), \(n=\frac{P V}{R T}=\frac{4.1 \times 5.86}{0.082 \times 293}=1 mol\)
