Question

The density of water vapour at 328.4 atm and 800 K is 135.0 g/dm'. Determine the molar volume, V.
and the compression factor of water vapour.​

Answer 1

Answer:

Volume is 0.2 L

Compressibility factor is 1.

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Answer 2

The molar volume, V of given water vapor is 7.5 L/mol and its compressibility factor is 37.5.

Explanation:

It is known that,

        Molar volume ($V_{m}$) = no. of moles (n) per liter of volume

As we know that,

               n = $\frac{mass}{\text{molar mass}}$

Hence,    $V_{m} = \frac{m}{M \times V}$ ......... (1)

Also,    density = $\frac{mass}{volume}$ ........ (2)

Thus, on putting the value of equation (2) into equation (1) the formula will be as follows.

            $V_{m} = \frac{density}{\text{molar mass}}$

Since, molar mass of water is 18 g/mol. Hence, the value of molar volume is as follows.

        Molar volume = $\frac{density}{\text{molar mass}}$

                                 = $\frac{135.0 g/dm^{3}}{18 g/mol}$

                                 = 7.5 L/mol

Now, formula to calculate the compressibility factor is as follows.

             Z = $\frac{PV_{m}}{RT}$

                 = $\frac{328.4 atm \times 7.5 L/mol}{0.0821 atm L/mol K \times 800 K}$

                 = 37.5

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