Problem 24 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The minimum work which must be done to compress \(16 \mathrm{~g}\) of oxygen isothermally, at \(300 \mathrm{~K}\) from a pressure of \(1.01325\) \(\times 10^{3} \mathrm{~N} / \mathrm{m}^{2}\) to \(1.01325 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\) is \((\ln 100=4.6, R=8.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol})\) (a) \(5727 \mathrm{~J}\) (b) \(11.454 \mathrm{~kJ}\) (c) \(123.255 \mathrm{~kJ}\) (d) \(1232.55 \mathrm{~J}\)

Step-by-Step Solution

Answer

5727 J

Step 1: Understanding the Exercise

The exercise is about calculating the minimum work required to isothermally compress oxygen. The initial and final pressures are given, as well as the temperature. The mass of oxygen and the gas constant are also provided. Use the formula for work done during isothermal compression for an ideal gas: \( W = nRT \ln\left(\frac{P_f}{P_i}\right) \)

Step 2: Calculate number of moles of oxygen

Given the mass of oxygen is 16 g and the molar mass of oxygen (\(O_2\) molecule) is 32 g/mol, calculate the number of moles (\(n\) by dividing the mass by the molar mass: \( n = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ mol} \).

Step 3: Calculate Isothermal Work

Using the formula for isothermal work, substitute the values for the number of moles (\(n\), gas constant (\(R\) equal to 8.3 J/K-mol), temperature (\(T\) which is 300 K), and the natural logarithm of the final pressure divided by the initial pressure (\(\ln(\frac{P_f}{P_i})\) which is given as 4.6): \( W = 0.5 \text{ mol} \times 8.3 \text{ J/K-mol} \times 300 \text{ K} \times 4.6 = 5727 \text{ J} \).

Key Concepts

thermodynamics

Thermodynamics is a fundamental branch of physics that deals with heat and temperature and their relation to energy and work. It comes down to understanding how different forms of energy can be transformed into heat or work and vice versa. The laws of thermodynamics govern these transformations and are critical in a widespread array of applications – from engines to living organisms, to atmospheric science. In the context of our exercise, it specifically pertains to the principles governing the isothermal (constant temperature) compression of gases. This is an example of a thermodynamic process where work is performed on a gas, resulting in a change in the gas's pressure and volume without a change in temperature.

physical chemistry

Physical chemistry uses the principles of physics to understand the microscopic properties and behavior of chemicals. It's a bridge between physics and chemistry, focusing on how matter is constructed at the molecular level and how it changes. Computations like the one in our exercise are a key aspect of physical chemistry, where the work done during a chemical process is predicted or understood. Physical chemists often use mathematical equations, such as the one for calculating work done on a gas during isothermal compression, to describe and predict the outcome of chemical processes at the molecular level.

gas laws

The gas laws are a series of fundamental principles that describe the behavior of gases and their interactions with changes in temperature, volume, and pressure. Boyle's Law, Charles's Law, and Gay-Lussac's Law are among the most well-known of these laws, each describing a different aspect of gas behavior. When these are combined, you get the ideal gas law, stated as PV=nRT, which is a cornerstone in understanding the behavior of ideal gases under various conditions. In our exercise, we refer to these laws implicitly when we use the formula for work done during isothermal compression. The formula reflects that the product of pressure and volume is constant during the process at a fixed temperature, indicating direct applications of the gas laws in thermodynamic processes.

work done by gas

Work done by a gas is a measure of the energy transferred when a gas expands or is compressed. When a gas is compressed, as in our textbook exercise, work is done on the gas. By convention, this work is considered positive, because energy is being transferred to the gas. For isothermal compression, the work is computed using the formula W = nRT ln(P_f/P_i), wherein n is the number of moles of gas, R is the gas constant, T is the absolute temperature, and P_f and P_i are the final and initial pressures respectively. The logarithmic relation indicates that the work depends not just on the initial and final states, but also on the path taken during the compression – and for an ideal gas during an isothermal process, that path follows a specific type of curve known as an isotherm.