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The Amazing World of Solutes, Solvents, and Solutions - Effect of pressure on density

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Density is defined as the mass of a substance per unit volume, mathematically expressed as ρ=mV\rho = \frac{m}{V}.

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In solids and liquids, the particles are already closely packed with very little intermolecular space. Therefore, applying pressure has a negligible (almost zero) effect on their volume and density.

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Gases are highly compressible because of the large intermolecular spaces between their particles.

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When pressure on a gas is increased at a constant temperature, the particles are pushed closer together, which reduces the volume (VV).

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Since density is inversely proportional to volume (ρ∝1V\rho \propto \frac{1}{V}) for a fixed mass, an increase in pressure leads to an increase in the density of the gas.

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In the context of solutions, pressure primarily affects the density of gaseous solutes or solutions involving gases (like carbonated water), while having little effect on liquid-liquid or solid-liquid solutions.

📐Formulae

Density(ρ)=Mass(m)Volume(V)\text{Density} (\rho) = \frac{\text{Mass} (m)}{\text{Volume} (V)}

P1V1=P2V2 (Boyle’s Law at constant temperature)P_1 V_1 = P_2 V_2 \text{ (Boyle's Law at constant temperature)}

ρ∝P (For a gas at constant temperature)\rho \propto P \text{ (For a gas at constant temperature)}

💡Examples

Problem 1:

A sample of oxygen gas has a mass of 32 g32\text{ g} and occupies a volume of 20 L20\text{ L} at a certain pressure. If the pressure is increased such that the volume reduces to 10 L10\text{ L}, calculate the initial and final density of the gas.

Solution:

Initial Density ρ1=32 g20 L=1.6 g/L\rho_1 = \frac{32\text{ g}}{20\text{ L}} = 1.6\text{ g/L}. Final Density ρ2=32 g10 L=3.2 g/L\rho_2 = \frac{32\text{ g}}{10\text{ L}} = 3.2\text{ g/L}.

Explanation:

Since the mass remains constant and the volume is halved due to increased pressure, the density doubles.

Problem 2:

Calculate the change in mass of a liquid solution if the pressure is increased from 1 atm1\text{ atm} to 10 atm10\text{ atm}, given the initial mass is 500 g500\text{ g}.

Solution:

500−5000\begin{array}{r} 500 \\ - 500 \\ \hline 0 \end{array}

Explanation:

Mass is an intrinsic property and does not change with pressure. Furthermore, for a liquid solution, the volume change is negligible, meaning the density remains almost constant despite the 9 atm9\text{ atm} increase.