
The total work done in a cyclic process is represented by the area enclosed by the graph in a PV diagram. The area can be calculated as the area of the rectangle formed by the points A, B, C, D, and E. The formula to find the work done is given by the area of the graph:
The change in pressure (height) is \( \Delta P = 400 \, \text{kPa} - 100 \, \text{kPa} = 300 \, \text{kPa} = 300 \times 10^3 \, \text{Pa} \).
The change in volume (width) is \( \Delta V = 4 \, \text{m}^3 - 2 \, \text{m}^3 = 2 \, \text{m}^3 \).
The area of a rectangle is \( \text{Area} = \text{height} \times \text{width} \).
Therefore, \( W = \Delta P \times \Delta V = (300 \times 10^3 \, \text{Pa}) \times (2 \, \text{m}^3) = 600 \times 10^3 \, \text{J} = 600 \, \text{kJ} \).
Corrected Calculation:
\(W = (400 - 100) \times 10^3 \, \text{Pa} \times (4 - 2) \, \text{m}^3\)
\(W = 300 \times 10^3 \, \text{Pa} \times 2 \, \text{m}^3\)
\(W = 600 \times 10^3 \, \text{J} = 600 \, \text{kJ}\)
Final Answer: The total work done is 600 kJ.
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 : \(A_2 \rightleftharpoons 2A\)