# 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? - Chemistry

Numerical

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?

#### Solution 1

Calculation of partial pressure of H2 in 1L vessel P1= 0.8 bar,

P2= ? V1= 0.5 L , V2 = 1.0 L

As temperature remains constant,

P1V1 = P2V2

(0.8 bar) (0.5 L) = P2 (1.0 L) or

P2 = 0.40 bar, i.e., PH2 = 0.40 bar

Calculation of partial pressure of 02 in 1 L vessel

P1‘ V1 = P2‘V2

(0.7 bar) (2.0 L) = P2 (1L) or

P2‘ = 1.4 bar, i.e.,Po2= 1.4 bar

Total pressure =PHz + PQ2

= 0.4 bar + 1.4 bar

= 1.8 bar

#### Solution 2

Let the partial pressure of H2 in the vessel be "P"_("H"_2)

Now

"p"_1 = 0.8 " bar"          "p"_2 = "p"_("H"_2) = ?

"V"_1 = 0.5L                  "V"_2 = 1 "L"

it is known that

"p"_1"V"_1 = "p"_2"V"_2

=> "p"_2 = ("p"_1"V"_1)/"V"_2

=> "p"_("H"_2)  = (0.8 xx 0.5)/1

= 0.4 bar

Now, let the partial pressure of O2 in the vessel be "po"_2

Now,

"p"_1 = 0.7 bar           "p"_2 = "po"_2 = ?

"V"_1 =  2.0 L             V_2 = 1 L

"p"_1"V"_1 = "p"_2"V"_2

⇒ "p"_2 = "p"_1"V"_1"V"_2

⇒ "pO"_2 = 0.7 × 2.01

= 1.4 bar

Total pressure of the gas mixture in the vessel can be obtained as:

"p"_"total" = "p"_("H"_2) + "po"_2

= 0.4 + 1.4

= 1.8 bar

Hence, the total pressure of the gaseous mixture in the vessel is 1.8 bar.

Is there an error in this question or solution?
Chapter 5: States of Matter - EXERCISES [Page 158]

#### APPEARS IN

NCERT Chemistry Part 1 and 2 Class 11
Chapter 5 States of Matter
EXERCISES | Q 5.8 | Page 158

Share