Question

What will be the pressure of the gaseous mixture when $0.5L$ of ${H_2}$ at $0.8bar$ and $2.0L$ of dioxygen at $0.7bar$ are introduced in a $1L$ vessel at ${27^ \circ }C$ ?

Answer 1

Hint: The ideal gas law states that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant. Ideal gas is a gas whose particles exhibit no attractive interactions whatsoever; at high temperatures and low pressures, gases behave close to ideally.
Formula used:
$PV = nRT$
$P = $ Pressure
$V = $ Volume
$n = $ Amount of substance
$R = $ Ideal gas constant
$T = $ Temperature

Complete answer:
Given:
${P_1} = 0.8bar$
${V_1} = 0.5L$
${P_2} = 0.7bar$
${V_2} = 2.0L$
To find: Pressure of gaseous mixture of ${H_2}$ and ${O_2}$
For ${H_2}$ gas,
Substituting the given values in the ideal gas formula,
${P_1}{V_1} = {n_1}RT$
$0.8 \times 0.5 = {n_{{H_2}}} \times 0.0821 \times 300$
${n_{{H_2}}} = 0.016$
For ${O_2}$ gas
Similarly, for ${O_2}$ , Substituting the given values in the ideal gas formula,
${P_2}{V_2} = {n_2}RT$
$0.7 \times 2 = {n_{{O_2}}}0.0821 \times 300$
${n_{{O_2}}} = 0.056$
For mixture
$PV = nRT$
$P \times 1 = ({n_{{H_2}}} + {n_{{O_2}}}) \times 0.0821 \times 300$
$P = (0.016 + 0.056) \times 0.0821 \times 300$
$P = 1.79bar$
Hence, the pressure of the gaseous mixture is $1.79bar$ .

Note:
When the particles of a gas are so far apart that they do not exert any attractive forces on one another, the gas is said to be ideal. There is no such thing as an ideal gas in real life, but at high temperatures and low pressures (conditions in which individual particles move very swiftly and are very far apart from one another, with essentially no interaction), gases behave very similarly to ideally.