ints the correction for particle volume. a for specific for each g as. a() is th

ID: 1033212 • Letter: I

Question

ints the correction for particle volume. a for specific for each g as. a() is the correction for in In this Section # Date Net pressure of Carbon Dioxide Gas trial 2 trial 1 trial 3 . The net pressure is found using the equation on page 42. We also have to account for the difference in height of the water: From page 42: With the height correction: P..-P,-Po where P, is the total pressure. pressure, ?,60 is the vapor pressure of water andP-ve-ra-ce where P, is the total pressure correction for the difference in the height of water inside and outside the cylinder Show the calculations for the net pressure for trial 1 below: Value of R Calculated from Experimental Data Using the Ideal Gas Law trial 1 trial 2 trial 3 Average Show the calculations for trial 1 below: use ideal gas law: PV nRT P in atmospheres, V in liters, n is in moles, R has units ofan Tis in Kelvin. Rearrange to:

Explanation / Answer

For trial 1

Pressure due to water height difference, PH cm H2O = 8.6

Pressure due to water height difference, PH mm Hg = 0.74 * 8.6 = 6.364

Water vapor pressure, PH2O, mm Hg = 16.5

Total pressure, PT mm Hg = 763.8

Pressure of gas, Pgas mm Hg = PT + PH - PH2O

= 753.664 mm Hg

Pressure of gas, Pgas atm = 753.664 / 760 = 0.992

Volume of gas collected = 204 mL = 0.204 L

Molar mass of NaHCO3 = 84 g/mol

Mass of NaHCO3 = 0.742 g

Moles of NaHCO3 = 0.742 / 84 = 0.00883

Temperature of gas, T = 19 °C = 292 K

Gas constant, R = P V / (n T)

= 0.992 atm * 0.204 L / (0.00883 mol * 292 K)

= 0.0784 L-atm/mol-K

Similarly for other trials we get results as shown in the table below:

Trial 1

Trial 2

Trial 3

PH, cm H2O

8.6

4.9

4.3

PH, mm Hg

6.364

3.626

3.182

PH2O, mm Hg

16.5

PT, mm Hg

763.8

V, mL

204

222

183

V, L

0.204

0.222

0.183

Pgas, mm Hg

753.664

750.926

750.482

Pgas, atm

0.992

0.988

0.987

Mass of NaHCO3, g

0.742

0.765

0.724

Moles of NaHCO3

0.00883

0.00911

0.00862

R, L-atm/mol-K

0.0784

0.0825

0.0718

Average R = (0.0784 + 0.0825 + 0.0718)/3 = 0.0776 L-atm/mol-K