Question

Consider water at 500 kPa and a specific volume of 0.2 m3/kg, what is the temperature (in oC)?

Answer 1

Answer:

$T=-272.9^{o}C$

Explanation:

We have the ideal gasses equation $PV=nRT$ and the expression for the specific volume $v=\frac{V}{m}$, that is the inverse of the density, and for definition the number of moles is equal to the mass over the molar mass, that is $n=\frac{m}{M}$

And we can relate the three equations as follows:

$PV=nRT$

Replacing the expression for n, we have:

$PV=\frac{m}{M}RT$

$P\frac{V}{m}=\frac{RT}{M}$

Replacing the expression for v, we have:

$Pv=\frac{RT}{M}$

Now resolving for T, we have:

$T=\frac{PvM}{R}$

Now, we should convert all the quantities to the same units:

-Convert 500kPa to atm

$500kPa*\frac{0.00986923}{1kPa}=4.93atm$

-Convert 0.2$\frac{m^{3}}{kg}$ to $\frac{L}{kg}$

$0.2\frac{m^{3} }{kg}*\frac{1L}{1m^{3}}=0.2\frac{L}{kg}$

- Convert the molar mass M of the water from $\frac{g}{mol}$ to $\frac{kg}{mol}$

$18\frac{g}{mol}=\frac{1kg}{1000g}=0.018\frac{kg}{mol}$

Finally we can replace the values:

$T=\frac{(4.93atm)(0.2\frac{L}{kg})(0.018\frac{kg}{mol})}{0.082\frac{atm.L}{mol.K}}$

$T=0.216K$

$T=0.216K-273.15\\T=-272.9^{o}C$