Question

Liquid sodium can be used as a heat transfer fluid in some nuclear reactors due to its high thermal conductivity and low neutron absorptivity. Its vapor pressure is 40.0 torr at 633 degree C and an enthalpy of vaporization of 1.00x10^2 kJ/mol. To what temperature can it be heated if a maximum of 500 torr pressure is allowed in the system?

Answer 1

1119.1 K

Step-by-step explanation:

From Clausius-Clapeyron equation:

`(dP)/(dT)=`Δ
`(h_(v) )/(R) ((1)/(T^(2) ) )dT`

The equation may be integrated considering the enthalpy of vaporization constant, and its result is:

`ln((P_(2) )/(P_(1) ) )=-`Δ
`(h_(v) )/(R)*((1)/(T_(2) )-(1)/(T_(1) ))`

Isolating the temperature
`T_(2)`

`T_(2)=(1)/((-R)/(dhv)*ln((P_(2) )/(P_(1))) +(1)/(T_(1)) )`

`T_(2)=(1)/((-8.314)/(1.00*10^5)*ln((500)/(40)) +(1)/(906.15))`

`T_(2)=1119.1K`

Note: Remember to change the units of the enthalpy vaporization to J/mol; and the temperatures must be in Kelvin units.

There is a format mistake with the enthalpy of vaporization, each 'Δ' correspond to that.