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

To solve this problem, we need to determine the ratio of work done by the gas to the heat absorbed when it changes from state A to state B on a \( V\text{-}T \) graph for a monatomic ideal gas.

1. Understanding the Graph:
   • The graph is a straight line in the \( V\text{-}T \) plane, indicating a linear relationship between volume \( V \) and temperature \( T \).
2. Work Done by the Gas:
   • For an ideal gas, work done is \( W = P \Delta V \).
   • From the linear \( V \propto T \) relation, the process is isobaric, so: \( W = nR \Delta T \).
3. Heat Absorbed by the Gas:
   • From the first law of thermodynamics: \( Q = \Delta U + W \).
   • For a monatomic ideal gas: \( \Delta U = \frac{3}{2} nR \Delta T \).
4. Ratio Calculation:
   • \( Q = \frac{3}{2} nR \Delta T + nR \Delta T = \frac{5}{2} nR \Delta T \).
   • \( \frac{W}{Q} = \frac{nR \Delta T}{\frac{5}{2} nR \Delta T} = \frac{2}{5} \).

Therefore, the required ratio is \( \frac{2}{5} \).