Can I have help with part B? in the calculation of Part (b) that are likely caus

Question

Can I have help with part B?

in the calculation of Part (b) that are likely causes of the discrepancy between the calculated and actual blood flows 6.52. The consticuent partial pressures of a gas in equilibriam with a liquid solution at 30FC and I atm containing 2 lb. SO/100 lba H:0 are p 31.6 mm Hg and po 176 mm Hg. The balance of the gas is air (a) Calculate the partial pressare of air. If you make any assumptions, state what they are (b) Suppose the only dara available on this sysem gave ps 176 mm Hg,but there was no information given on the oquilibrium partial pressure of water. Use Raoult's law to estimate a valae for this quancity Assuming that the value given in the problem statement is correct, what pereentage eror results from using Raouk's law? (e) The same system was examined in Example 6.4-1. What percentage errors in the two calculated quantities would result from using Raoult's law for the partial pressure of water? 6.53. The solubility coefficient of a gas may be defined as the number of cubic centimeters (STP) of the gas that dissolves in l cm, of a solvent under a partial pressure of I alm·The solubility coefficient of CO2 (a) Calculate the Henry's law constant in atms'mole fraction for CO in H,O at 20°C from the given solubelity coefficient. (b) How many grams of CO, can be dissolved in a 12-oz bocle of soda at 20'C if the gas above the

Explanation / Answer

Solution:

(a) Given, the solubility coefficient of CO2 in water at 20oC = 0.0901 cm3/cm3 H2O/atm.

At STP, 22.4 L CO2 = 1 mol CO2

            22.4x103 cm3 CO2 = 1 mol CO2

                        0.0901 cm3 CO2 = 0.0901/22.4x103 = 4x10-6 mol CO2

Also, let density of water at 20oC = 1 g/cm3

Therefore, 1 cm3 H2O = 1 g H2O = 1/18 mol H2O

According to Henry’s law we know: Cg = kPg -------------(1)

Where, Cg and Pg are the concentration of the dissolved gas and its partial pressure. k = Henry’s solubility constant/coefficient.

Using equation (1): Cg = kPg

= 0.0901 (cm3/cm3 H2O/atm) x(1 atm)                      [since, Pg = 1 atm]

= 4x10-6 mol CO2/1g H2O

= 4x10-6x18 mol CO2/mol H2O

Or, kPg = 4x10-6x18 mol CO2/mol H2O = 72x10-6 mol CO2/mol H2O

            = 7.2x10-5 mol CO2/mol H2O

Or, k = 7.2x10-5 mol CO2/mol H2O/atm                      [since, Pg = 1 atm]

Now, mole fraction of CO2 = 7.2x10-5/(1+7.2x10-5) = 7.19x10-5

Therefore, k = 7.19x10-5 mole fraction of CO2/atm

(b) some part of the question is not available to answer part-b.

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