9. If a gas is stored at 2 atm pressure in 5.0Lcontainer, what will be the press

Question

9. If a gas is stored at 2 atm pressure in 5.0Lcontainer, what will be the pressure of gas if it is transferred to a 7.0L container? hat will be the pressure of a gas in a 250 ml can on a hot summer day, when temperature outside is 40 °C, if its pressure in winter when temperature was 20° C was recorded as 2.5 atm. from 22 °C to 27°C. Find the enthalpy of solution in ki/mol for the reaction. NaOH (s) 11. When solid NaOH (2. 5 g ) was dissolved in 100. ml of water, temperature of water changed (aq) MICROTECQ 800-634-2718 9:33 P 11/13/20 ENG

Explanation / Answer

Ans. #9. Using Ideal gas equation:            PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature (in K) = (0C + 273.15) K

# Putting the values in equation 1 for case 1 (2 atm, 5.0 L)

            PV = n1 RT                                                     - R, T remains constant

            Or, 2.0 atm x 5.0 L = n1 RT

            Or, n1 = 10.0 atm L / RT

# Putting the values in equation 1 for case 1 (7.0 L)

Change in V, P, and T does not affect the number of moles of the gas. So, the number of moles of gas remains constant at n1.

            PV = n1 RT                                                     - R, T remains constant

            Or, P x 7.0 L = (10.0 atm L / RT) RT = 10.0 atm L

            Or, P = 10.0 atm L / 7.0 L = 1.43 atm

Therefore, pressure of gas in 7.0 L container = 1.43 atm

#10. Putting the values in equation 1 for case 1 (2.5 atm, 20.00C)

            2.5 atm x 0.250 L = n x 0.0821 atm L mol-1K-1 x 293.15 K

            Or, n = 0.625 atm L / 24.067615 atm L mol-1 = 0.026 mol

Hence, moles of gas = 0.026 mol

Change in V, P, and T does not affect the number of moles of the gas. So, the number of moles of gas remains constant at 0.026 mol.

# Now, using equation 1 for case 2 (40.00C)

            P x 0.250 L = 0.026 mol x 0.0821 atm L mol-1K-1 x 313.15 K

            Or, P = 0.66844999 atm L / 0.250 L = 2.67 atm

Hence, pressure at 40.00C = 2.67 atm

#11. Heat gained by water during solvation of NaOH is given by-

q = m s dT                            - equation 1

Where,

q = heat gained

m = mass

s = specific heat

dT = Final temperature – Initial temperature

Putting the values in equation 1-

q = 100.0 g x (4.184 J g-10C-1) x (27.0 – 22.0)0C

Or, q = 2092.0 J

# The total amount of heat gained by the water to increase its temperature must be equal to the total amount of heat released during dissolution of NaOH.

So,

            Heat released by 2.5 g NaOH dissolution = -2092.0 J

Note: the –ve sign indicates release of energy during solation of NaOH.

# Moles of NaOH solvated = Mass of NaOH / Molar mass

                                                = 2.5 g / (40.0 g/mol)

                                                = 0.0625 mol

# Now,

Molar enthalpy of NaOH solution = Heat released / Moles of NaOH solvated

                                                = -2092.0 J / 0.0625 mol

                                                = -33472 J/ mol

                                                = -33.472 kJ/mol

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