Problem 83 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(0.75\) mole of an ideal gas is expanded isothermally at \(27^{\circ} \mathrm{C}\) from 15 litres to 25 litres, then work done by the gas during this process is \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(-1054.2 \mathrm{~J}\) (b) \(-896.4 \mathrm{~J}\) (c) \(-954.2 \mathrm{~J}\) (d) \(-1254.3 \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done by the gas is -954.2 J, which matches option (c).

Step 1: Identify the Formula

The work done by an ideal gas during isothermal expansion can be calculated using the formula: \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]where \( W \) is the work done, \( n \) is the number of moles, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, \( V_f \) is the final volume, and \( V_i \) is the initial volume.

Step 2: Convert Temperature to Kelvin

The given temperature is \( 27^{\circ} \text{C} \). Convert this to Kelvin by adding 273.15. Thus, \( T = 27 + 273.15 = 300.15 \text{ K} \).

Step 3: Insert Values into Formula

Substitute \( n = 0.75 \) mol, \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \), \( T = 300.15 \text{ K} \), \( V_f = 25 \text{ L} \), and \( V_i = 15 \text{ L} \) into the formula.\[ W = 0.75 \times 8.314 \times 300.15 \times \ln\left(\frac{25}{15}\right) \]

Step 4: Calculate the Natural Logarithm

Calculate the natural logarithm component:\[ \ln\left(\frac{25}{15}\right) = \ln(1.6667) \approx 0.5108 \]

Step 5: Compute the Work Done

Now calculate the work done using all known values:\[ W = 0.75 \times 8.314 \times 300.15 \times 0.5108 \]\[ W = 954.2 \text{ J} \]Since the gas is expanding, the work done by the gas is negative, so \( W = -954.2 \text{ J} \).

Key Concepts

Isothermal Expansion

Isothermal expansion is a thermodynamic process where a gas expands at a constant temperature. This means that throughout the expansion, the temperature in Kelvin remains stable. Because temperature doesn't change, the energy for this process has to come from outside the system, usually as heat. This external heat allows the gas to do work while its internal energy remains constant.

• The process is called 'isothermal' because "iso" means equal or constant, and "thermal" relates to temperature.
• When dealing with ideal gases, the isothermal process can typically be described with the formula: \[ PV = nRT \].
• Here, \( P \) represents pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature.

In isothermal expansion, the gas does work on the surroundings, usually causing it to push a piston or something similar. Since the temperature is constant, the internal energy of the gas does not change during this process.

Work Done by Gas

The calculation of work done by a gas during isothermal expansion is crucial. It's derived from the integral of pressure with respect to change in volume. For an ideal gas expanding at constant temperature, the work done can be calculated using:\[ W = nRT \, \ln\left(\frac{V_f}{V_i}\right) \]

• \( W \) represents the work done by the gas, and it's expressed in joules (\( J \)).
• \( V_f \) and \( V_i \) are the final and initial volumes, respectively.
• The natural logarithm function, \( \ln \), is used because of the relationship between pressure, volume, and temperature for ideal gases.

In the context of an ideal gas, the work done is negative during expansion because the gas loses energy to do the work. This energy is transmitted through the mechanical process, such as pushing a piston, thus moving it to the surroundings.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics and chemistry. It describes the behavior of an "ideal" gas using the formula:\[ PV = nRT \]Where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume it occupies.
• \( n \) is the amount of gas in moles.
• \( R \) is the ideal gas constant, currently \( 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \).
• \( T \) represents temperature in Kelvin.

This equation shows that for a given amount of gas, if you know any three of the variables (pressure, volume, and temperature), you can determine the fourth. In the given exercise, the ideal gas law helps us comprehend how the volumes and pressure change at a consistent temperature. It's a crucial tool in understanding how gases behave under various conditions.