An ideal gas is taken around the cycle ABCA as shown in P-V diagram below :

The work done by the gas during the cycle is equal to

ANS:C - 3 P₁V₁

To find the work done by the gas during the cycle, we need to calculate the area enclosed by the cycle on the P-V diagram. Given that the cycle is ABCA, it forms a closed loop on the P-V diagram. The work done by the gas during a cycle on a P-V diagram is equal to the area enclosed by the cycle. This can be calculated using the formula: Work=∮𝑃𝑑𝑉Work=∮PdV where 𝑃P is pressure and 𝑑𝑉dV is the change in volume. Looking at the diagram, we see that the cycle ABCA consists of two sections:

1. A to B: Isobaric process (constant pressure)
2. B to C: Isochoric process (constant volume)

For the A to B process, the work done is given by: Work𝐴→𝐵=𝑃1⋅(𝑉2−𝑉1)WorkA→B​=P1​⋅(V2​−V1​) For the B to C process, since it's an isochoric process (constant volume), the work done is zero. Therefore, the total work done during the cycle is equal to the work done in the A to B process: Total Work=Work𝐴→𝐵=𝑃1⋅(𝑉2−𝑉1)Total Work=WorkA→B​=P1​⋅(V2​−V1​) Given that 𝑉2=2𝑉1V2​=2V1​ (as per the diagram), we can substitute this into the equation: Total Work=𝑃1⋅(2𝑉1−𝑉1)=𝑃1⋅𝑉1Total Work=P1​⋅(2V1​−V1​)=P1​⋅V1​ So, the work done by the gas during the cycle is 𝑃1⋅𝑉1P1​⋅V1​. Therefore, the correct option is 𝑃1𝑉1P1​V1​.