Question

Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically work done is (A) more in an isothermal process (B) more in an adiabatic process (C) equal in both process. (D) neither of them

Short answer

The correct answer is (B) more in an adiabatic process. This is because \(W_{isothermal} = nRT \ln(\frac{V_f}{V_i})\) and \(W_{adiabatic} = \frac{nRT(V_f - V_i)}{\gamma - 1}\), and since \( \ln(\frac{V_f}{V_i}) < \frac{V_f - V_i}{\gamma - 1}\), we can conclude that \(W_{isothermal} < W_{adiabatic}\).

Step-by-step solution

Understand the Concepts of Isothermal and Adiabatic Processes

An isothermal process is a process that occurs at a constant temperature. In an isothermal expansion, the gas expands while the temperature remains constant by absorbing heat from its surroundings. The work done for an ideal gas in an isothermal process is given by: \( W_{isothermal} = nRT \ln(\frac{V_f}{V_i}) \) Where \(W_{isothermal}\) is the work done during the isothermal process, n is the number of moles, R is the gas constant, T is the temperature, and \(V_f\) and \(V_i\) are the final and initial volumes, respectively. An adiabatic process is a process that occurs without any exchange of heat between the system and its surroundings. In an adiabatic expansion, the gas expands without gaining or losing any heat from its surroundings. The work done for an ideal gas in an adiabatic process is given by: \( W_{adiabatic} = \frac{nRT(V_f - V_i)}{\gamma - 1} \) Where \(W_{adiabatic}\) is the work done during the adiabatic process and γ (gamma) is the heat capacity ratio.

Identify the Constraints

Since we are given two identical samples of gas expanding, we know that the number of moles (n), the constant temperature (T), and the initial and final volumes (\(V_i\) and \(V_f\)) will be the same in both processes. Therefore, the factors determining the work done are the coefficients nRT and γ (gamma) for the isothermal and adiabatic processes, respectively.

Compare the Work Done in Both Processes

We now need to compare the work done in isothermal, \(W_{isothermal}\), and adiabatic processes, \(W_{adiabatic}\), using the equations discussed in Step 1: \( W_{isothermal} = nRT \ln(\frac{V_f}{V_i}) \) \( W_{adiabatic} = \frac{nRT(V_f - V_i)}{\gamma - 1} \) Considering that the logarithmic function, ln(x), is less than x for all x > 1, we can say: \( \ln(\frac{V_f}{V_i}) < \frac{V_f - V_i}{\gamma - 1} \) Since nRT is positive, multiplying both sides of the inequality by nRT does not change the direction of the inequality: \( W_{isothermal} < W_{adiabatic} \) Therefore, more work is done in the adiabatic process compared to the isothermal process.

Determine the Correct Answer

Based on our comparison in Step 3, we can conclude that the correct answer is: (B) more in an adiabatic process