Question

The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

• A. \(\frac{2}{5}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{2}{7}\)

Answer 1

The Correct Option is A

The problem requires determining the ratio of the work done by a monatomic gas to the heat absorbed by it as it undergoes a change from state A to state B shown in the PV diagram. Let's analyze this step-by-step.

1. The given graph shows the volume (V) vs temperature (T) for a monatomic gas. We can assume an ideal gas where the state equation is given by \( PV = nRT \).
2. First Law of Thermodynamics: The change in internal energy \( \Delta U \) of the gas is given by: \[\Delta U = Q - W\] where \( Q \) is the heat absorbed and \( W \) is the work done by the gas.
3. Monatomic Gas Internal Energy Change: The change in internal energy for a monatomic ideal gas is: \[\Delta U = \frac{3}{2} nR\Delta T\] Here, \( \Delta T \) is the change in temperature.
4. Work Done (W): For a process at constant pressure, work done can be expressed as: \[W = P \Delta V = nR \Delta T\] due to the ideal gas law.
5. Ratio Calculation: The problem asks for the ratio of work done by the gas to the heat absorbed: \[\frac{W}{Q} = \frac{nR \Delta T}{Q}\] Using the first law, rearrange to express \( Q \): \[Q = \Delta U + W = \frac{3}{2}nR\Delta T + nR\Delta T = \frac{5}{2}nR\Delta T\] Therefore, the ratio becomes: \[\frac{W}{Q} = \frac{nR \Delta T}{\frac{5}{2}nR \Delta T} = \frac{2}{5}\]
6. The correct option is hence found to be: \(\frac{2}{5}\), which matches the given correct answer.