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Sphinxa [80]
3 years ago
7

Calculate the height of a column of liquid benzene (d=0.879g/cm3), in meters, required to exert a pressure of 0.790 atm .

Chemistry
2 answers:
Simora [160]3 years ago
7 0

Here we have to get the height of the column in meter, filled with liquid benzene which exerting pressure of 0.790 atm.

The height of the column will be 0.928 m.

We know the relation between pressure and height of a liquid placed in a column is: pressure (P) = Height (h) × density of the liquid (ρ) × gravitational constant (g).

Here the pressure (P) is 0.790 atm,

or [0.790 × (1.013 × 10⁶)] dyne/cm². [As 1 atm is equivalent to 1.013 × 10⁶ dyne/cm²]

Or, 8.002ₓ10⁵ dyne/cm².

density of benzene is given 0.879 g/cm³.

And gravitational constant (g) is 980 cm/sec².

On plugging the values we get:

8.002×10⁵ = h × 0.879 × 980

Or, h = 928.931 cm

Or, h = 9.28 m (As 1 m = 100 cm)

Thus the height will be 9.28 m.

OverLord2011 [107]3 years ago
3 0

Answer:

9.29x10^-3m

Explanation:

Step 1:

Data obtained from the question. This includes:

Density (ρ) = 0.879g/cm3

Pressure (P) = 0.790 atm

Height (h) =?

Step2:

Conversion to appropriate units. This is very vital in order to obtain the answer to the question in the desired unit.

For density:

1g/cm3 = 1x10^6g/m3

Therefore, 0.879g/cm3 = 0.879x1x10^6 = 879000g/m3

For pressure:

1 atm = 101325N/m2

Therefore, 0.790 atm = 0.790x101325 = 80046.75N/m2

Step 3:

Determination of the height

Pressure is related to height according to the following equation:

Pressure = Density x acceleration due to gravity x height

P = ρgh

P = 80046.75N/m2

ρ = 879000g/m3

g = 9.8m/s2

h =?

80046.75 = 879000 x 9.8 x h

Divide both side by 879000 x 9.8

h = 80046.75 / (879000 x 9.8)

h = 9.29x10^-3m

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For an isothermal process equation will be as follows.

                W = nRT ln\frac{P_{1}}{P_{2}}

It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.

                    No. of moles = \frac{mass}{\text{molar mass}}

                                           = \frac{10000 g/s}{18 g/mol}

                                           = 555.55 mol/s

                                           = 556 mol/s (approx)

As T = 50^{o}C or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.

                  W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]

                      = 556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}    

                     = 556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303    

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                     3440193.809 J/s \times \frac{0.001 kJ}{1 J}

                          = 3440.193 kJ/s

                          = 3451 kJ/s (approx)

Thus, we can conclude that the pump work is 3451 kJ/s.

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