Question

An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is _____ $ \times 10^{-1} $ J. (Take $ \pi = 3.14 $)

Hint:

The work done in a cyclic process is the area enclosed by the path on a \( PV \)-diagram. For a circular path, use the area formula \( A = \pi r^2 \) to determine the work done by the gas.

Answer 1

Correct Answer: 31.4

The graph is a circle in the PV diagram, and for a cyclic process, the work done by the gas is equal to the area enclosed by the cycle.

• Radius in pressure: \( R_P = \frac{500 - 300}{2} = 100\, \text{kPa} \)
• Radius in volume: \( R_V = \frac{350 - 150}{2} = 100\, \text{cm}^3 \)

Assuming both axes are scaled equally, the radius \( R = 100 \) \[ \text{Area} = \pi R^2 = 3.14 \times 100^2 = 3.14 \times 10^4 \] Convert units: \[ 1\, \text{kPa} \cdot \text{cm}^3 = 10^{-2} \, \text{J} \Rightarrow \text{Work done} = 3.14 \times 10^4 \times 10^{-2} = 314 \, \text{J} \] 
Work done = $31.4 \times 10^{-1}$ J

Answer 2

Given the area of the circle \( W = \frac{\pi}{4} d_1 d_2 \), we can compute the work done as: \[ W = \frac{\pi}{4} (500 - 300) \times 10^3 \times (350 - 150) \times 10^{-6} \] Simplifying: \[ W = 31.4 \, \text{Joule} \] Thus: \[ W = 314 \times 10^{-1} \, \text{Joule} \] \[ \boxed{W = 31.4 \, \text{Joule}} \]