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

Answer:

1119.1 K

Explanation:

From Clausius-Clapeyron equation:

$\frac{dP}{dT}=$Δ$\frac{h_{v} }{R} (\frac{1}{T^{2} } )dT$

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

$ln(\frac{P_{2} }{P_{1} } )=-$Δ$\frac{h_{v} }{R}*(\frac{1}{T_{2} }-\frac{1}{T_{1} })$

Isolating the temperature $T_{2}$

$T_{2}=\frac{1}{\frac{-R}{dhv}*ln(\frac{P_{2} }{P_{1}}) +\frac{1}{T_{1}} }$

$T_{2}=\frac{1}{\frac{-8.314}{1.00*10^5}*ln(\frac{500}{40}) +\frac{1}{906.15}}$

$T_{2}=1119.1K$

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.