Explanation:
The given data is as follows.
No. of moles = [tex]1.70 \times 10^{-3}[/tex], V = ?
T = 20.2 + 273 K = 293.2 K, P = [tex]\frac{795 mm Hg}{760.0 mm Hg/atm}[/tex] = 1.046 atm, R = [tex]0.0821 L atm K^{-1}mol ^{-1}[/tex]
Calculate the volume using ideal gas equation as follows.
P V = n R T
[tex]1.046 atm \times V = 1.70 \times 10^{-3} \times 0.0821 L atm K^{-1}mol ^{-1} \times 293.2 K[/tex]
V = [tex]\frac{1.70 \times 10^{-3} \times 0.0821 L atm K^{-1}mol ^{-1} \times 293.2 K}{1.046 atm}[/tex]
= [tex]\frac{40.921 L atm}{1.046 atm}[/tex]
= [tex]39.122 \times 10^{-3}[/tex] L
Thus, we can conclude that volume of the gas is [tex]39.122 \times 10^{-3}[/tex] L.
