What will be the pressure of the gaseous mixture when 0.5 L of H, at 0.8 bar and 2 L of oxygen at 0.7 bar are introduced in a 1 L vessel at $27^{\circ} C$ ?

Open in App

Solution

We know that
$P V = n R T$
$\Rightarrow n = \frac{P V}{R T}$ .
mole of $H_{2} = \frac{P V}{R T}$
For $H_{2}$ , pressure $(P) = 0.8$ bar & $V = 0.5 L$
moles of $H_{2} = \frac{0.8 \times 0.5}{R T} = \frac{0.4}{R T}$ .
For $O_{2}$ $P = 0.7$ bar, Volume $(V) = 2 L$
moles of $O_{2} = \frac{0.7 \times 2}{R T} = \frac{1.4}{R T}$
Total number of moles= $\frac{0.4}{R T} + \frac{1.4}{R T} = \frac{1.8}{R T}$
The total pressure= $\frac{1.8}{R T} \times \frac{R T}{1} = 1.8 a t m$

Suggest Corrections

Similar questions

Q. What will be the pressure of the gaseous mixture when

0.5 L

H_{2}

0.8 b a r

and

2.0 L

O_{2}

0.7 b a r

are introduced in a

1 L

vessel at

27^{\circ} C

What will be the pressure of the gaseous mixture when 0.5 L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?

Q. What will be the pressure of the gaseous mixture when

0.5 L

H_{2}

0.8 b a r

and

2 L

of dioxygen at

0.7 b a r

are introduced in a

1 L

vessel at

27^{\circ} C

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 $27^{\circ} C$ ?

Join BYJU'S Learning Program

What will be the pressure of the gaseous mixture when 0.5 L of H, at 0.8 bar and 2 L of oxygen at 0.7 bar are introduced in a 1 L vessel at 27∘C?

We know thatPV=nRT⇒n=PVRT.mole of H2=PVRTFor H2, pressure (P)=0.8 bar & V=0.5Lmoles of H2=0.8×0.5RT=0.4RT .For O2 P=0.7 bar, Volume (V)=2Lmoles of O2=0.7×2RT=1.4RTTotal number of moles= 0.4RT+1.4RT=1.8RTThe total pressure= 1.8RT×RT1=1.8atm

What will be the pressure of the gaseous mixture when 0.5 L of H, at 0.8 bar and 2 L of oxygen at 0.7 bar are introduced in a 1 L vessel at $27^{\circ} C$ ?