chemistry
CHE 242 Tutorial 3: Problems & Answers Week 4: Feb 2- 6, 2009
1. Consider a gas that occupies 1.00 dm3 at a pressure of 2.00 bar. If the gas is compressed isothermally at constant external pressure, Pext, so that the final volume is 0.500 dm3, what is the smallest value Pext can have? Calculate the work involved using this value of Pext.
Solution
For compression to occur, the value of Pext must be at least as large as the final pressure of the gas. P1 = 2.00 bar, V1 = 1.00 dm3, V2 = 0.500 dm3, P2 = ? = 4.00 bar
This is the smallest value of Pext can be to compress the gas isothermally (constant temperature) from 1.00 dm3 to 0.500 dm3. Note: 1 bar = 105 Pa
= 2.00 dm3 bar = (2.00 dm3 bar)(10-3 m dm-3)(105 Pa bar-1) = 200 Pa m3 = 200 J OR (in SI units)
(4.00 x 105 Pa){(0.500 – 1.00) x 10-3 m3) = 200 J
Of course, Pext can be any value greater than 4.00 bar, so 200 J represents the smallest value of w for the isothermal compression at constant pressure from a volume of 1.00 dm3 to 0.500 dm3.
2. A sample of CO2 (g) occupy 2.00 dm3 mol-1 at a temperature of 300 K. If the gas is compressed isothermally at a constant external pressure, Pext, so that the final molar volume is 0.750 dm3 mol-1, calculate the smallest value Pext can have, assuming that CO2 (g) satisfies the van der Waals equation of state under these conditions. Calculate the work done on the gas using this value of Pext.
Solution
P1 = 2.00 dm3, T1 = T2 = 300K (isothermal process) , = 2.00 dm3, = 0.750 dm3 van der Waals equation:
For CO2, a = 3.6551 dm6 bar mol-2, b = 0.042816 dm3 mol-1
The smallest value Pext can have is P2, where P2 is the final pressure of the gas. = 28.8 bar
-(28.8 x 105 Pa){(0.750 – 2.00) x 10-3 m3} = 3600 J = 3.60 kJ
3. (a) Calculate the work involved when 2 moles of an ideal gas are compressed reversibly from 1.00 bar to 5.00 bar at a constant temperature of 300 K.
(b) Calculate the work done if instead the
1. Consider a gas that occupies 1.00 dm3 at a pressure of 2.00 bar. If the gas is compressed isothermally at constant external pressure, Pext, so that the final volume is 0.500 dm3, what is the smallest value Pext can have? Calculate the work involved using this value of Pext.
Solution
For compression to occur, the value of Pext must be at least as large as the final pressure of the gas. P1 = 2.00 bar, V1 = 1.00 dm3, V2 = 0.500 dm3, P2 = ? = 4.00 bar
This is the smallest value of Pext can be to compress the gas isothermally (constant temperature) from 1.00 dm3 to 0.500 dm3. Note: 1 bar = 105 Pa
= 2.00 dm3 bar = (2.00 dm3 bar)(10-3 m dm-3)(105 Pa bar-1) = 200 Pa m3 = 200 J OR (in SI units)
(4.00 x 105 Pa){(0.500 – 1.00) x 10-3 m3) = 200 J
Of course, Pext can be any value greater than 4.00 bar, so 200 J represents the smallest value of w for the isothermal compression at constant pressure from a volume of 1.00 dm3 to 0.500 dm3.
2. A sample of CO2 (g) occupy 2.00 dm3 mol-1 at a temperature of 300 K. If the gas is compressed isothermally at a constant external pressure, Pext, so that the final molar volume is 0.750 dm3 mol-1, calculate the smallest value Pext can have, assuming that CO2 (g) satisfies the van der Waals equation of state under these conditions. Calculate the work done on the gas using this value of Pext.
Solution
P1 = 2.00 dm3, T1 = T2 = 300K (isothermal process) , = 2.00 dm3, = 0.750 dm3 van der Waals equation:
For CO2, a = 3.6551 dm6 bar mol-2, b = 0.042816 dm3 mol-1
The smallest value Pext can have is P2, where P2 is the final pressure of the gas. = 28.8 bar
-(28.8 x 105 Pa){(0.750 – 2.00) x 10-3 m3} = 3600 J = 3.60 kJ
3. (a) Calculate the work involved when 2 moles of an ideal gas are compressed reversibly from 1.00 bar to 5.00 bar at a constant temperature of 300 K.
(b) Calculate the work done if instead the