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. $\frac{2}{7}$
• B. $\frac{2}{5}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The Correct Option is B

To find the ratio of the work done by the gas to the heat absorbed when the gas changes from state \(A\) to state \(B\), we need to analyze the behavior of a monatomic ideal gas under these conditions.

1. Understanding the Process: The volume \(V\) of a monatomic ideal gas changes with its temperature \(T\). Based on the provided information (implied from the graph), assume this is a process where pressure remains constant (isobaric process), as no specific information about the graph is provided here.
2. Work Done by the Gas: In an isobaric process, the work done by the gas is given by: \[ W = P \Delta V \] where \(P\) is the constant pressure and \(\Delta V\) is the change in volume.
3. Heat Absorbed by the Gas: For a monatomic ideal gas undergoing an isobaric process, the heat absorbed \(Q\) is given by: \[ Q = nC_P \Delta T \] where \(n\) is the number of moles, \(C_P\) is the molar heat capacity at constant pressure, and \(\Delta T\) is the change in temperature.
4. Relating Work and Heat: The work done during an isobaric process can also be related to temperature change as: \[ W = nR \Delta T \] where \(R\) is the ideal gas constant.
5. Finding the Ratio: The ratio of work done to the heat absorbed is: \[ \frac{W}{Q} = \frac{nR \Delta T}{nC_P \Delta T} = \frac{R}{C_P} \] For a monatomic ideal gas, \(C_P = \frac{5}{2}R\), thus: \[ \frac{W}{Q} = \frac{R}{\frac{5}{2}R} = \frac{2}{5} \]

Hence, the ratio of work done by the gas to the heat absorbed is \(\frac{2}{5}\).