Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $${\tau _1}$$ to $${\tau _2}$$ . If $${{{C_p}} \over {{C_v}}} = \gamma $$ for this gas then a good estimate for $${{{\tau _2}} \over {{\tau _1}}}$$ is given by :

Two moles of an ideal gas with $${{{C_P}} \over {{C_V}}} = {5 \over 3}$$ are mixed with 3 moles of another ideal gas with $${{{C_P}} \over {{C_V}}} = {4 \over 3}$$. The value of $${{{C_P}} \over {{C_V}}}$$ for the mixture is :

A litre of dry air at STP expands adiabatically to a volume of 3 litres. If $$\gamma $$ = 1.40, the work done by air is : (3^1.4 = 4.6555) [Take air to be an ideal gas]

One kg of water, at 20^oC, heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20 $$\Omega $$. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to : [Specific heat of water = 4200 J/(kg ^oC), Latent heat of water = 2260 kJ/kg]