The average kinetic energy (\( KE \)) of gas molecules is directly proportional to the absolute temperature (\( T \)) and is given by: \[ KE = \frac{3}{2} k T, \] where \( k \) is Boltzmann's constant. \bigskip
Step 1: Initial kinetic energy at \( 27^\circ \text{C} \). The absolute temperature at \( 27^\circ \text{C} \) is: \[ T_1 = 27 + 273 = 300 \, \text{K}. \]
Step 2: New temperature for double kinetic energy. If the kinetic energy becomes double: \[ KE_2 = 2 \cdot KE_1. \] Since \( KE \propto T \): \[ T_2 = 2 \cdot T_1 = 2 \cdot 300 = 600 \, \text{K}. \] Convert \( T_2 \) to Celsius: \[ T_2 = 600 - 273 = 327^\circ \text{C}. \]
Final Answer: The temperature is: \[ \boxed{327^\circ \text{C}}. \]
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,

The correct relation between $\gamma=\frac{ c _{ p }}{ c _{ v }}$ and temperature $T$ is :
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 : \(A_2 \rightleftharpoons 2A\)