Question

At a constant temperature, 250 mL of nitrogen at 760 mm pressure and 500 mL of oxygen at 600 mm pressure are put together in a one litre flask. What will be the partial pressure of two gases and the total pressure of mixture of the above non reacting gases in one litre flask

Answer 1

Answer:

P

1

V

1

+P

2

V

2

=P

mix

V

mix

(At const. temperature)

⇒(760)(250)+(500)(600)=(1000)P

mix

⇒P

mix

=300+190=490mm=Final Pressure

Answer 2

Answer: The partial pressure of nitrogen is 296 mm Hg and that of oxygen is 366 mm Hg. The total pressure is 662 mm Hg.

Explanation:  

According to ideal gas equation :

$PV=nRT$

$\n_{N_2}=\frac{p_{N_2}\times V_{N_2}}{RT}=\frac{760\times 250}{RT}$

$\n_{O_2}=\frac{p_{O_2}\times V_{O_2}}{RT}=\frac{600\times 500}{RT}$

$x_{O_2}=\frac{n_{O_2}}{n_{N_2}+n_{O_2}}=\frac{\frac{600\times 500}{RT}}{\frac{760\times 250}{RT}+\frac{600\times 500}{RT}}=0.61$

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_A=x_A\times p_A^0$ and  $p_B=x_B\times p_B^0$

and  

where, x = mole fraction

$p^0$ = pressure in the pure state

$p_{N_2}=0.39\times 760=296 mm Hg$

$p_{O_2}=0.61\times 600=366 mm Hg$

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_{N_2}+p_{O_2}=296+366=662mm Hg$