The magnitude of heat exchanged by a system for the given cyclic process ABC (as shown in the figure) is (in SI units):
In thermodynamics, the heat exchanged by a system in a cyclic process is equal to the area enclosed by the process curve on a \( P-V \) diagram. In the given problem, the process involves a rectangle on the \( P-V \) diagram (since the pressure-volume graph forms a closed loop between points A, B, and C). The area of this rectangle can be calculated as: \[ \text{Area} = \text{Length} \times \text{Width} = (400 - 200) \times (200 - 100) = 200 \times 100 = 10\pi \text{ (in appropriate units)}. \] Therefore, the magnitude of heat exchanged is \( 10\pi \) units.
Final Answer: \( 10\pi \).
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).