Question

$120 \,g$ of an ideal gas of molecular weight $40\, g \,mol ^{-1}$ are confined to a volume of $20 \,L$ at $400\, K$. Using $R=0.0821 \,L$ atm $K^{-1} \,mol ^{-1}$, the pressure of the gas is

• A. 4.90 atm
• B. 4.92 atm
• C. 5.02 atm
• D. 4.96 atm

Answer 1

Answer: (B)

$P V=R T, $
$P V=\frac{W}{M} R T$
$20 P=\frac{120}{40} \times .0821 \times 400$
or $P=4.92 \,atm$