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PilotLPTM [1.2K]
3 years ago
6

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?
Physics
1 answer:
Leya [2.2K]3 years ago
5 0

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.

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