Question

A box contains 0.90 g of liquid water in equilibrium with water vapour at 27°C. The equilibrium vapour pressure of water at 27°C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If the liquid water evaporates, then the volume of the box must be - litre (nearest integer) R = 0.0821 L atm K^-1 mol^-1.
(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas)

पर्याय

• 14 L

• 55 L

• 29 L

• 10 L

Answer 1

29 L

Explanation:

Since, 760 Torr = 1 atm

∴ 32 Torr = `32/760` atm

As all the liquid water evaporates so entire water is in gaseous state.

Weight of water vapour = 0.9 g

∴ Moles of water vapour (n) = `"Given mass"/"Molecular mass" = 0.9/18`

Pressure (P) = `32/760` atm

Temperature (T) = (27 + 273) K = 300 K

R = 0.082 L atm K^-1 mol^-1

Given water vapour acts as an ideal gas, so we can apply ideal gas equation.

From ideal gas equation,

PV = n RT

`32/760 xx "V" = 0.9/18 xx 0.082 xx 300`

V = `(0.9 xx 0.082 xx 300 xx 760)/(32 xx 18)`

V = 29.21 L

V = 29 L (nearest integer)