Question

A closed system consists of \(0.5 \mathrm{kmol}\) of liquid water and occupies a volume of \(4 \times 10^{-3} \mathrm{~m}^{3}\). Determine the weight of the system, in \(\mathrm{N}\), and the average density, in \(\mathrm{kg} / \mathrm{m}^{3}\), at a location where the acceleration of gravity is \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

Short answer

Weight: 88.29 N, Density: 2250 kg/m\textsuperscript{3}

Step-by-step solution

Determine the Weight of the System

First, find the molar mass of water. The molar mass of water (H\textsubscript{2}O) is approximately \(18 \text{g/mol} = 0.018 \text{kg/mol} \). Next, calculate the mass of the water using the given amount of substance:\[ M = 0.5 \text{kmol} \times 0.018 \text{kg/mol} = 9 \text{kg} \].Now use the gravity to find the weight: \[ W = M \times g = 9 \text{kg} \times 9.81 \text{m/s}^2 = 88.29 \text{N} \].

Calculate the Average Density

Use the given volume and the calculated mass to find the average density. The formula for density\( \rho \) is\[ \rho = \frac{M}{V} \].Using the mass \( M = 9 \text{kg} \) and the volume \( V = 4 \times 10^{-3} \text{m}^{3} \), \[ \rho = \frac{9 \text{kg}}{4 \times 10^{-3} \text{m}^{3}} = 2250 \text{kg/m}^{3} \].

Key concepts

Closed System

In thermodynamics, a **closed system** refers to a system where mass is conserved within the boundaries. This means that no mass enters or leaves the system, even though energy transfer (like heat and work) can occur. For example, think of a sealed bottle of water. No matter how much you shake or heat it, the amount of water inside stays the same. The exercise we looked at involves such a closed system, with a specific amount of liquid water trapped within a fixed volume.

Density Calculation

Density is a measure of how much mass exists in a given volume. To calculate density (\text{\rho}), use the formula: \text{\[ \rho = \frac{M}{V} \]}. Here, \text{\rho} is the density, \text{M} is the mass, and \text{V} is the volume. In the given exercise, the mass of water is 9 kg, and the volume is \text{4 \times 10^{-3}} m\text{^{3}}. Substituting these values into the formula, we get: \text{\[ \rho = \frac{9 \text{kg}}{4 \times 10^{-3} \text{m}^{3}} = 2250 \text{kg/m}^3 \]}. This value indicates how much mass is present per unit volume, which is essential for understanding the properties of the system.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol) or kilograms per mole (kg/mol). It provides a bridge between macroscopic and molecular-scale measurements. For water (H\textsubscript{2}O), the molar mass is approximately \text{18 g/mol}, or \text{0.018 kg/mol}. This value helps in converting the amount of substance (in moles) to mass (in kilograms). For example, in the exercise, \text{0.5 kmol} of water needs to be converted to mass. By multiplying the amount of substance by the molar mass (\text{\[ M = 0.5 \text{kmol} \times 0.018 \text{kg/mol} = 9 \text{kg} \]}), we determine the mass of water in the system.

Weight

Weight is the force exerted by gravity on an object's mass. It can be calculated using the formula: \text{\[ W = M \times g \]}. Here, \text{W} is weight, \text{M} is mass, and \text{g} is the acceleration due to gravity (approximately \text{9.81 m/s\text{^{2}}}). In the exercise, the mass of water is \text{9 kg}. Using the formula, we get: \text{\[ W = 9 \text{kg} \times 9.81 \text{m/s}^{2} = 88.29 \text{N} \]}. This illustrates how gravity acts on the mass of water, providing the force (weight) of the closed system.