We are given that the gas obeys the equation PV³ = RT during the path from A to B, and the process is cyclic.
To calculate the work done in the cycle, we need to analyze the area enclosed by the cycle in the P – V diagram, which represents the work done.
The work done in a cyclic process is given by the area enclosed by the cycle on the P – V diagram. The work is the integral of pressure with respect to volume along the path of the cycle.
Since the gas obeys the equation PV³ = RT, the work done can be calculated by finding the area under the curve from A to B and then calculating the work done along the other parts of the cycle.
By calculating the areas based on the graph provided, the total net work done over the complete cycle is found to be:
Wtotal = 205 J.
Thus, the net work done in the complete cycle is 205 J, and the correct answer is Option (2).
The left and right compartments of a thermally isolated container of length $L$ are separated by a thermally conducting, movable piston of area $A$. The left and right compartments are filled with $\frac{3}{2}$ and 1 moles of an ideal gas, respectively. In the left compartment the piston is attached by a spring with spring constant $k$ and natural length $\frac{2L}{5}$. In thermodynamic equilibrium, the piston is at a distance $\frac{L}{2}$ from the left and right edges of the container as shown in the figure. Under the above conditions, if the pressure in the right compartment is $P = \frac{kL}{A} \alpha$, then the value of $\alpha$ is ____
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).