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 $) 
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.
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
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: \[ W = 314 \times 10^{-1} \, \text{Joule} \] \[ \boxed{W = 31.4 \, \text{Joule}} \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)