Laboratory 1: Ideal Gas Law

Instructor: Barry Thomas

Course: PHY 202-11

Report By: Eric Diaz

Partner:

Felicia Freeman

Data Received: January 20, 2011

Objective:

The objective of this lab is to prove experimentally that air behaves like an ideal gas. Boyle’s Law will be used to try and acquire constant values, which would imply that the behavior of air obeys the Ideal Gas Law.

Data:

Data of the Apparatus:

Diameter = D = 3.5 mm

Cross-sectional Area of the Tube = A = 9.62 x 10-2 cm2

Xt = 82 cm + 6.3 cm = 88.3cm

Data of the Environment:

Atmospheric Pressure = Pa = 73.2 Hg∙cm

Temperature = 297 K

Part 1: When V and T are constant.

Case Starting Point Open Tube Close TubeCase 1 X = 35 cm Xo = 47.7 cm Xc = 40 cmCase 2 X = 30 cm Xo = 55.3 cm Xc = 40 cmCase 3 X = 25 cm Xo = 62.6 cm Xc = 40 cmCase 4 X = 20 cm Xo = 70.2 cm Xc = 40 cm

Part 2: When n and T are constant.

Starting Point for case 1 and 2 = Xc = Xo = 30cm

Case 1

Close Tube Open TubeXc = 31 cm Xo = 32.3 cmXc = 32 cm Xo = 34.8 cmXc = 33 cm Xo = 37.0 cmXc = 34 cm Xo = 39.8 cmXc = 35 cm Xo = 42.1 cmXc = 36 cm Xo = 44.7 cmXc = 37 cm Xo = 47.2 cmXc = 38 cm Xo = 50.3 cmXc = 39 cm Xo = 52.7 cmXc = 40 cm Xo = 55.5 cm

Case 2

Close Tube Open TubeXc = 29 cm Xo = 27.7 cmXc = 28 cm Xo = 25.5 cmXc = 27 cm Xo = 23.5 cmXc = 26 cm Xo = 21.3 cmXc = 25 cm Xo = 19.2 cmXc = 24 cm Xo = 17.1 cmXc = 23 cm Xo = 15.1 cmXc = 22 cm Xo = 13.2 cmXc = 21 cm Xo = 11.1 cmXc = 20 cm Xo = 9.20 cm

Calculations:

Part 1: Case 1

Starting point X = 35 cm

VA = (Xt – 35 cm) A

VA = (88.3 cm – 35 cm) (9.62 x10-2 cm2) = 5.13 cm3

P = PA + ( Xo – Xc )

P = 73.2 Hg∙ cm + ( 47.7 cm – 40 cm ) = 80.9 Hg∙ cm

ρ = (4.645 x 10-3) ( PAT ) ρ = (4.645 x 10-3) ( 73.2

297 ) = 1.14 x 10-3 gm/cm3

m = ρVA

m = (1.14 x 10-3 gm/cm3 )(5.13 cm3) = 5.85 x 10-3gm

n = mM

n = 5.85E−3 gm29 gm /mol = 2.02 x 10-4 mol

Pn

= Constant

80.9Hg∙cm2.02E−4mol

= 4.00 x 105 Hg∙cm/mol

Part 1: Case 2

Starting point X = 30 cm

VA = (Xt – 30 cm) A

VA = (88.3 cm – 30 cm) (9.62 x10-2 cm2) = 5.61 cm3

P = PA + ( Xo – Xc )

P = 73.2 Hg∙ cm + ( 55.3 cm – 40 cm ) = 88.5 Hg∙cm

ρ = (4.645 x 10-3) ( PAT ) ρ = (4.645 x 10-3) ( 73.2

297 ) = 1.14 x 10-3 gm/cm3

m = ρVA

m = (1.14 x 10-3 gm/cm3 )(5.61 cm3) = 6.40 x 10-3gm

n = mM

n = 6.40 E−3 gm29 gm /mol = 2.21 x 10-4 mol

Pn

= Constant

88.5Hg∙cm2.21E−4mol

= 4.00 x 105 Hg∙cm/mol

Part 1: Case 3

Starting point X = 25 cm

VA = (Xt – 25 cm) A

VA = (88.3 cm – 25 cm) (9.62 x10-2 cm2) = 6.09 cm3

P = PA + ( Xo – Xc )

P = 73.2 Hg∙ cm + ( 62.6 cm – 40 cm ) = 95.8 Hg∙cm

ρ = (4.645 x 10-3) ( PAT ) ρ = (4.645 x 10-3) ( 73.2

297 ) = 1.14 x 10-3 gm/cm3

m = ρVA

m = (1.14 x 10-3 gm/cm3 )(6.09 cm3) = 6.94 x 10-3gm

n = mM

n = 6.94 E−3gm29 gm /mol = 2.39 x 10-4 mol

Pn

= Constant

95.8Hg∙cm2.39E−4mol

= 4.01 x 105 Hg∙cm/mol

Part 1: Case 4

Starting point X = 20 cm

VA = (Xt – 20 cm) A

VA = (88.3 cm – 20 cm) (9.62 x10-2 cm2) = 6.57 cm3

P = PA + ( Xo – Xc )

P = 73.2 Hg∙ cm + ( 70.2 cm – 40 cm ) = 103 Hg∙cm

ρ = (4.645 x 10-3) ( PAT ) ρ = (4.645 x 10-3) ( 73.2

297 ) = 1.14 x 10-3 gm/cm3

m = ρVA

m = (1.14 x 10-3 gm/cm3 )(6.57 cm3) = 7.49 x 10-3gm

n = mM

n = 6.94 E−3gm29 gm /mol = 2.58 x 10-4 mol

Pn

= Constant

103Hg∙cm2.58E−4mol

= 3.99 x 105 Hg∙cm/mol

Part 2: Case 1

P = PA + ( Xo – Xc ) VA = (Xt – Xo cm) A

1. P = 73.2 + (32.3 – 31) = 74.5 Hg∙ cm VA = (88.3 cm – 31 cm) (9.62E-2 cm2) = 5.51cm3

2. P = 73.2 + (34.8 – 32) = 76 Hg∙ cm VA = (88.3 cm – 32cm) (9.62E-2 cm2) = 5.42 cm3

3. P = 73.2 + (37.0 – 33) = 77.2 Hg∙ cm VA = (88.3 cm – 33 cm) (9.62E-2 cm2) = 5.32 cm3

4. P = 73.2 + (39.8 – 34) = 79 Hg∙ cm VA = (88.3 cm – 34 cm) (9.62E-2 cm2) = 5.22 cm3

5. P = 73.2 + (42.1 – 35) = 80.3 Hg∙ cm VA = (88.3 cm – 35cm) (9.62E-2 cm2) = 5.13 cm3

6. P = 73.2 + (44.7 – 36) = 81.9 Hg∙ cm VA = (88.3 cm – 36cm) (9.62E-2 cm2) = 5.03 cm3

7. P = 73.2 + (47.2 – 37) = 83.4 Hg∙ cm VA = (88.3 cm – 37 cm) (9.62E-2 cm2) = 4.94 cm3

8. P = 73.2 + (50.3 – 38) = 85.5 Hg∙ cm VA = (88.3 cm – 38 cm) (9.62E-2 cm2) = 4.84 cm3

9. P = 73.2 + (52.7 – 39) = 86.9 Hg∙ cm VA = (88.3 cm – 39 cm) (9.62E-2 cm2) = 4.74 cm3

10. P = 73.2 + (55.5 – 40) = 88.7 Hg∙ cm VA = (88.3 cm – 40 cm) (9.62E-2 cm2) = 4.65 cm3

PV = Constant

PV= (74.5)(5.51) = 4.10 x 102 J

PV= (76)(5.42) = 4.12 x 102 J

PV= (77.2)(5.32) = 4.11 x 102 J

PV= (79)(5.22) = 4.12 x 102 J

PV= (80.3)(5.13) = 4.12 x 102 J

PV= (81.9)(5.03) = 4.12 x 102 J

PV= (83.4)(4.94) = 4.12 x 102 J

PV= (85.5)(4.84) = 4.14 x 102 J

PV= (86.9)(4.74) = 4.12 x 102 J

PV= (88.7)(4.65) = 4.12 x 102 J

Part 2: Case 2

P = PA + ( Xo – Xc ) VA = (Xt – Xo cm) A

1. P = 73.2 + (27.7 – 29) = 71.9Hg∙ cm VA = (88.3 cm – 29 cm) (9.62E-2 cm2) = 5.70cm3

2. P = 73.2 + (25.5 – 28) = 70.7 Hg∙ cm VA = (88.3 cm – 28cm) (9.62E-2 cm2) = 5.80 cm3

3. P = 73.2 + (23.5 – 27) = 69.7 Hg∙ cm VA = (88.3 cm – 27 cm) (9.62E-2 cm2) = 5.90 cm3

4. P = 73.2 + (21.3 – 26) = 68.5 Hg∙ cm VA = (88.3 cm – 26 cm) (9.62E-2 cm2) = 5.99 cm3

5. P = 73.2 + (19.2 – 25) = 67.4 Hg∙ cm VA = (88.3 cm – 25cm) (9.62E-2 cm2) = 6.09 cm3

6. P = 73.2 + (17.1 – 24) = 66.3 Hg∙ cm VA = (88.3 cm – 24cm) (9.62E-2 cm2) = 6.19 cm3

7. P = 73.2 + (15.1 – 23) = 65.3 Hg∙ cm VA = (88.3 cm – 23 cm) (9.62E-2 cm2) = 6.28 cm3

8. P = 73.2 + (13.2 – 22) = 64.4 Hg∙ cm VA = (88.3 cm – 22 cm) (9.62E-2 cm2) = 6.38cm3

9. P = 73.2 + (11.1 – 21) = 63.3 Hg∙ cm VA = (88.3 cm – 21 cm) (9.62E-2 cm2) = 6.47 cm3

10. P = 73.2 + (9.2 – 20) = 62.4 Hg∙ cm VA = (88.3 cm – 20 cm) (9.62E-2 cm2) = 6.57 cm3

PV = Constant

PV= (71.9)(5.70) = 4.10 x 102 J

PV= (70.7)(5.80) = 4.10 x 102 J

PV= (69.7)(5.90) = 4.11 x 102 J

PV= (68.5)(5.99) = 4.10 x 102 J

PV= (67.4)(6.09) = 4.10 x 102 J

PV= (66.3)(6.19) = 4.10 x 102 J

PV= (65.3)(6.28) = 4.10 x 102 J

PV= (64.4)(6.38) = 4.11 x 102 J

PV= (63.3)(6.47) = 4.10 x 102 J

PV= (62.4)(6.57) = 4.10 x 102 J

Question:

Does air behave like an ideal gas, at least at room temperature?

Yes, air does behave like an ideal gas. The results seem to be very consistent, which allow the conclusion that PV = nRT could be applied to air and acquire reliable results.

Results:

Part 1:

Case 1: 4.00 x 105 Hg∙cm/mol

Case 2: 4.00 x 105 Hg∙cm/mol

Case 3: 4.01 x 105 Hg∙cm/mol

Case 4: 3.99 x 105 Hg∙cm/mol

Part 2:

Case 1: Case 2:

PV4.10 x 102 J4.12 x 102 J4.11 x 102 J4.12 x 102 J4.12 x 102 J4.12 x 102 J4.12 x 102 J4.14 x 102 J4.12 x 102 J4.12 x 102 J

PV4.10 x 102 J4.10 x 102 J4.11 x 102 J4.10 x 102 J4.10 x 102 J4.10 x 102 J4.10 x 102 J4.11 x 102 J4.10 x 102 J4.10 x 102 J