Question

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

Answer 1

`T=-272.9^(o)C`

Step-by-step explanation:

We have the ideal gasses equation
`PV=nRT` and the expression for the specific volume
`v=(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=(m)/(M)`

And we can relate the three equations as follows:

`PV=nRT`

Replacing the expression for n, we have:

`PV=(m)/(M)RT`

`P(V)/(m)=(RT)/(M)`

Replacing the expression for v, we have:

`Pv=(RT)/(M)`

Now resolving for T, we have:

`T=(PvM)/(R)`

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

-Convert 500kPa to atm

`500kPa*(0.00986923)/(1kPa)=4.93atm`

-Convert 0.2
`(m^(3))/(kg)` to
`(L)/(kg)`

`0.2(m^(3) )/(kg)*(1L)/(1m^(3))=0.2(L)/(kg)`

- Convert the molar mass M of the water from
`(g)/(mol)` to
`(kg)/(mol)`

`18(g)/(mol)=(1kg)/(1000g)=0.018(kg)/(mol)`

Finally we can replace the values:

`T=((4.93atm)(0.2(L)/(kg))(0.018(kg)/(mol)))/(0.082(atm.L)/(mol.K))`

`T=0.216K`

`T=0.216K-273.15\\T=-272.9^(o)C`