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: \[ \boxed{327^\circ \text{C}}. \]
0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion, produced 0.9 g H₂O. Molar mass of (X) is ___________g mol\(^{-1}\).
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to: