The volume occupied by 1.8 g of water vapour at 374 °C and 1 bar pressure will be :- [Use R = 0.083 bar L K-1mol-1]

Question

The volume occupied by 1.8 g of water vapour at 374 °C and 1 bar pressure will be :- [Use R = 0.083 bar L K-1mol-1] (A) 96.66 L (B) 55.87 L (C) 3.10 L (D) 5.37 L

Tardigrade Admin · Accepted Answer

PV = nRT n = 1.8/18 = 0.1 mole P = 1 bar T = 374 + 273 = 647 K V = (nRT/P) = (0.1 × 0.083 × 647/1) = 5.37 L

Q. The volume occupied by 1.8 g of water vapour at 374 °C and 1 bar pressure will be :-
[Use R = 0.083 bar L $K^{- 1} m o l^{- 1}$ ]

96.66 L

55.87 L

3.10 L

5.37 L

$P V = n RT$
$n = 1.8/18 = 0.1$ mole
$P = 1$ bar
$T = 374 + 273$
$= 647 K$
$V = \frac{n RT}{P}$
$= \frac{0.1 \times 0.083 \times 647}{1}$
$= 5.37 L$

Q. The volume occupied by 1.8 g of water vapour at 374 °C and 1 bar pressure will be :- [Use R = 0.083 bar L K−1mol−1]