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 1L vessel at o 27°C ?

Asked by Pragya Singh | 1 year ago |  57

1 Answer

Solution :-

Let the partial pressure of \( {H_{ 2 }}\)​ in the container be \( p_{H_{ 2 }}\)​​.

Now,

\( {p_{ 1 }}\) = 0.8 bar

\( {p_{ 2 }} = p_{H_{ 2 }} \)

\( {V_{ 1 }} = 0.5 L\)

\( {V_{ 2 }}= 1 L\)

It is known that,

\( {p_{ 1 }}​ {V_{ 1 }}= {p_{ 2 }}​ {V_{ 2 }}\)

\( {p_{ 2 }}= \dfrac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }} p_{H_{ 2 }}\)

\( = \dfrac{ 0.8 \; \times \; 0.5 }{ 1 }\)

= 0.4 bar

Now, let the partial pressure of O2 in the container be \( p_{O_{ 2 }}\).

Now,

\( {p_{ 1 }}= 0.7bar\)

\( {V_{ 1 }}= 2.0 L\)

\( {V_{ 2 }}= 1 L\)

\( {p_{ 2 }}​ = p_{O_{ 2 }}\) = ?

\( {p_{ 1 }}{V_{ 1 }}​ = {p_{ 2 }}{V_{ 2 }}\)

\( {p_{ 2 }}= \dfrac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }}\)

\( p_{O_{ 2 }} = \dfrac{ 0.7 \; \times \; 20 }{ 1 }\)

= 1.4 bar

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

\( p_{total}= p_{H_{ 2 }} + p_{O_{ 2 }}\)

= 0.4 + 1.4

= 1.8 bar

Answered by Abhisek | 1 year ago

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