Question

If inside a container of volume $8.21$ litres water vapours are introduce at a rate of $0.1atm$ per sec then how many moles of water gets liquified at $300K$ after $9$ seconds (Vapour pressure of water $=0.3atm$ at $300K)$.

• A. $0,1$
• B. $0.2$
• C. $0.3$
• D. None of these

Answer 1

Answer: (A)

$0,1$

Given,

Volume of a gas=$8.21L$

Temperature=$300K$

Rate=$0.1atm/s$

After 9 seconds, the pressure exerted on gas=$0.1atm/s\times 9s$

$\Rightarrow P=0.9atm$ after 9 seconds.

Thus, initially, the pressure exerted on the gas will be equal to vapour pressure since vapour pressure is the pressure exerted by a gas in equilibrium with the condensed phase.

The number of moles presents initially can be calculated by using an ideal gas equation.

$PV=nRT$

Where,

P-pressure exerted on gas

V-volume

n-number of moles

R-gas constant ($0.08206atm{K}^{-1}mo{l}^{-1}$ )

T-temperature

$\Rightarrow 0.3atm\times 8.21L=n\times 0.08206atm{K}^{-1}mo{l}^{-1}\times 300K$

$\Rightarrow \frac{2.463}{24.62}mol=n$

$n=0.1mol$

Thus, $0.1mol$ is present at vapour pressure of a gas.

From Boyle's law, it is clear that at a constant temperature, the pressure and volume are constant.

$P_1{V}_1=P_2{V}_2$ .......(1)

where,

$P_1{V}_1$- pressure and volume initially (vapour pressure)

$P_2{V}_2$-pressure and volume after 9 seconds

From ideal gas equation, it is clear that $PV=nRT$

Substitute ideal gas equation in equation 1, we get

$n_{1}RT=n_{2}RT$ .......(2)

Since temperature is constant

we know that $0.1mol$ is present in a gas of volume $8.21L$ at $0.3atm$ i.e)$n_1=0.1mol$

$\left(2\right)\Rightarrow n_1=n_2$

Thus, the number of moles at $0.9atm$ will be equal to the number of moles of a gas at vapour pressure (i.e $0.1mol$)

$n_2=0.1mol$

Thus, option-A is correct.