
To solve the problem, we need to understand the characteristics of adiabatic and isothermal processes depicted in the \( P-V \) diagram.
In the diagram, two adiabatic paths intersect with two isothermal curves. The points \(a, b, c,\) and \(d\) represent specific volume and pressure conditions. We are given four volume values: \(V_a, V_d, V_b,\) and \(V_c\).
Concept Explanation:
Reasoning:
For adiabatic processes between the same isothermals, the relations are such that:
Thus, the correct relation is: \(\frac{V_a}{V_d} = \frac{V_b}{V_c}\).
For an adiabatic process, the equation \(TV^{\gamma-1} = \text{constant}\) holds.
Between points \(a\) and \(d\):
\[ T_a \cdot V_a^{\gamma-1} = T_d \cdot V_d^{\gamma-1}. \]
\[ \frac{V_a}{V_d} = \frac{T_d}{T_a}. \]
Between points \(b\) and \(c\):
\[ T_b \cdot V_b^{\gamma-1} = T_c \cdot V_c^{\gamma-1}. \]
\[ \frac{V_b}{V_c} = \frac{T_c}{T_b}. \]
Given \(T_d = T_c\) and \(T_a = T_b\), we have:
\[ \frac{V_a}{V_d} = \frac{V_b}{V_c}. \]
Final Answer: \(\frac{V_a}{V_d} = \frac{V_b}{V_c}\).
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 : \(A_2 \rightleftharpoons 2A\)