#### Question

An amount *n* (in moles) of a monatomic gas at an initial temperature T_{0} is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature T* _{s}* (> T

_{0}) and the atmospheric pressure is P

_{α}. Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area A, thickness

*x*and thermal conductivity K. Assuming all changes to be slow, find the distance moved by the piston in time

*t*.

#### Solution

In time *dt**,* heat transfer through the bottom of the cylinder is given by

`"dQ"/"dt" = "KA(T_s - T_0)"/x`

For a monoatomic gas, pressure remains constant.

∴ `dQ = nC_pdT`

∴ `(nC_pdT)/ dt = "KA(T_2 - T_0)"/x`

For a monoatomic gas,

`C_p = 5/2 R`

`⇒ "n5RdT"/"2dt" = KA(T_s - T_0)/x`

`⇒ "5nR"/2 "dT"/dt = (KA(t_s - T_0))/x`

`⇒ "dT"/(T_s - T_0) = "-2KAdt"/"5nRx"`

Integrating both the sides,

`(T_s - T_0)_"T_0"^"T" = "-2KAt"/"5nRx"`

`⇒ In ((T_s - T) /(T_s - T_0)) = - "-2KAt"/"5nRx"`

`⇒ T_s - T = (T_s - T_0)e ^("-2KAt"/"5nRx")`

`⇒ T = T_s - (T_s - T_0) =e ^(-"-2KAt"/"5nRx")`

`⇒ T - T_0 = (T_s - T_0) - (T_s - T_0)e^(-"2KAt"/"5nRx"`

`⇒ T- T_0 = (T_s - _0) [l - e^(-"-2KAt"/"5nRx")]`

From the gas equation,

`(P_(a)Al)/(nR) = T - T_0`

∴ `(P_(a)Al)/(nR)= (T_s - T_0) [1 - e^(-"-2KAt"/"5nRx")]`

`⇒ l = (nR)/(P_aA) (T_s - T_0)[ 1 - e^(-"-2KAt"/"5nRx")]`