One mode of an ideal monatomic gas is kept in a rigid vessel. The vessel is kept inside a steam chamber whose temperature is `97^(@)C` . Initially, the temperature of the gas is `5.0^(@)C` . The walls of the vessel have an inner surface of area `800cm^(2)` and thickness `1.0cm` If the temperature of the gas increases to `9.0^(@)C` in `0.5` second, find the thermal conductivity of the material of the walls.

Explanation

The intial temperature difference is `97^(@)C-5^(@)C=92^(@)C` and at `5.0s` the temperature difference become `97^(@)C-9^(@)C=88^(@)C` . As the change in the temperature difference is small, we work with the average temperature difference
`(92^(@)C+88^(@)C)/2=90^(@)C=90K` . The rise in the temperature of the gas is
`9.0^(@)C-5.0^(@)C=4^(@)C=4K` . The heat supplied to the gas in `5.0s` is
`DeltaQ=nC_(v)DeltaT` . `=(1mol)xx((3)/(2)xx8.3JK^(-1)mol^(-1))xx(4K)` . `=49.8J` . If the thermal conductivity is K, `49.8J=(K(800xx10^(_4)m)^(2)xx(90K))/(1.0xx10^(-2)m)xx5.0s` . or, `K=(49.08J)/(3600msK)=0.014Js^(-1)m^(-1)K^(-1)` .