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

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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?

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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.

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Chapter 5: States of Matter - EXERCISES [Page 158]

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

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