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bulgar [2K]
3 years ago
14

On a warm summer day in Oslo, the temperature is 21 degree C and the relative humidity is 70%. What is the absolute humidity? Wh

at is the humid volume of the air? What is the specific enthalpy of the air, taking H_2O (L, 0 degree C, 1 atm) as a reference state? What is the wet bulb temperature? What is the dew point temperature?
Chemistry
1 answer:
lesya [120]3 years ago
4 0

Explanation:

According to the psychrometric chart at dry bulb temperature of 21^{o}C and RH 70%, the absolute humidity = 0.011 kg/kg dry air

Formula to calculate humid volume is as follows.

       \nu_{H} m^{3}/kg dry air = \frac{22.41}{273}TK (\frac{1}{28.97} + \frac{1}{18.02}H)

                     = (2.83 \times 10^{-3} + 4.56 \times 10^{-3} H) T K

                     = (2.83 \times 10^{-3} + 4.56 \times 10^{-3} \times 0.011) \times 294

                       = 0.847 m^{3}/kg dry air

Hence, humid volume of air is 0.847 m^{3}/kg dry air.

Specific enthalpy of dry air  = specific heat capacity of dry air × dry bulb temperature

                    = 1.006 kJ/kg ^{o}C \times 21^{o}C

                    = 21.126 kJ/kg dry air

Hence, the specific enthalpy of the air is 21.126 kJ/kg dry air.

As per the psychrometric chart at given conditions wet bulb temperature = 17.5 ^{o}C

As per the psychrometric chart at given conditions dew point temperature = 15.5 ^{o}C

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Answer: The molecular of the compound is, C_2H_3O

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where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

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In 44 g of carbon dioxide, 12 g of carbon is contained.

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For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.902g of water, \frac{2}{18}\times 1.092=0.121g of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (1.621)-[(0.844)+(0.121)]=0.656g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.041 moles.

For Carbon = \frac{0.0703}{0.041}=1.71\approx 2

For Hydrogen  = \frac{0.121}{0.041}=2.95\approx 3

For Oxygen  = \frac{0.041}{0.041}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is C_2H_3O_1=C_2H_3O

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n=\frac{46.06}{43}=1

Molecular formula = (C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O

Therefore, the molecular of the compound is, C_2H_3O

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