Question

What will be the pressure of the gaseous mixture when 0.5 L of $H_{2}$ at 0. 8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C ?

Answer 1

Partial pressure of Hydrogen ${ H }_{ 2 }$,

${ V }_{ 1 }$ = 0.5 L,

${ V }_{ 2 }$ = 1 L

${ P }_{ 1 }$ = 0.8 bar

Boyle’s Law ${ { P }_{ 1 }V }_{ 1 }={ P }_{ 2 }{ V }_{ 2 }$

Therefore, ${ P }_{ 2 }=$ 0.8×0.5=0.4 bar

Partial pressure of dioxygen ${ O }_{ 2 }$,

${ V }_{ 1 }$ = 2 L,

${ V }_{ 2 }$ = 1 L,

${ P }_{ 1 }$ = 0.7 bar

Boyle’s Law ${ { P }_{ 1 }V }_{ 1 }={ P }_{ 2 }{ V }_{ 2 }$

Therefore, ${ P }_{ 2 }$ = 2×0.7 = 1.4 bar

P = ${ P }_{ { H }_{ 2 } }{ P }_{ O_{ 2 } }=$ 0.4 + 1.4 = 1.8 bar.

Answer 2

Answer:

(i) The partial pressure of hydrogen gas

P

′

=

V

′

PV

=

1

0.8×0.5

=0.40 bar

(ii) The partial pressure of oxygen gas

P

′′

=

V

′

PV

=

1

0.7×2.0

=1.40 bar

So, the total pressure of gaseous mixture is P+P

′′

=0.40+1.40=1.80 bar