Question

125 mL of a gas A of pressure 500 mm is mixed with 200 mL of another gas B at a pressure of 300 mm in a vessel of 150 mL capacity. What will be the total pressure of the resulting mixture if the temperature is kept constant?

Answer 1

Hint: Boyle’s law states that when a gas is held at a constant temperature and mass in a closed container, the volume and pressure vary inversely. The equation to use in this is
${p_1}{v_1} = {p_2}{v_2}$

Complete step by step answer:
The gasses behave as ideal gasses and won’t condense or react. All the gas is transferred to the final 150 ml container. The pressure is in mm. Each gas can be treated separately.
We should note that PV = nRT and thus PV is constant.
So, the equation is ${p_1}{v_1} = {p_2}{v_2}$
Given in the question are:
${V_1}$ = Volume of gas A = 125 mL
${V_2}$= Volume of gas B = 200 mL
${P_1}$= Pressure of gas A = 500 mm
${P_2}$= Pressure of gas B = 300 mm
Applying mole conservation,
${P_{total}} = \dfrac{{{P_1}{V_1} + \:{P_2}{V_2}}}{{{V_{vessel}}}}$
Substituting the values in the equation, we get,
 ${P_{total}} = \dfrac{{125\: \times 500 + 200\: \times 300}}{{150}}$​
$= 816.67$ mm

Note: Boyle’s law, which is also known as Mariotte’s law, is a relation which concerns the compression and expansion of a gas at constant temperature. This empirical relation, which was formulated by the physicist Robert Boyle in 1662, states that the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature; that is, in equation form, pv = k, a constant. The relationship was also discovered by the French physicist Edme Mariotte (1676). Real gases obey Boyle’s law at sufficiently low pressures, although the product pv generally decreases slightly at higher pressures, where the gas begins to depart from ideal behaviour.