8/29/2010 1 AUGUST 27 TH 2010 Chem 251 TUT + SOLUTIONS 1.5 A constant Volume perfect gas thermometer indicates a pressure of 6.69 kPa at the triple point temperature of water (273.16 K). (a) what change of pressure indicates a change of 1.00 K at this temperature? (b) What pressure indicates a temperature of 100.00C? (C) what change of pressure indicates a change of 1.00 K at the latter temperature Solutions Solutions Solutions Solutions a) a) a) a) This is a perfect gas, thus can use P This is a perfect gas, thus can use P This is a perfect gas, thus can use P This is a perfect gas, thus can use P 1 V 1 /T /T /T /T 1 = P = P = P = P 2 V 2 /T /T /T /T 2 Since this is at constant volume; equation simplifies to: P Since this is at constant volume; equation simplifies to: P Since this is at constant volume; equation simplifies to: P Since this is at constant volume; equation simplifies to: P 1 /T /T /T /T 1 = P = P = P = P 2 /T /T /T /T 2 T 2 = the change + original temperature = 1.00 k + 273.16 = 274.16 k = the change + original temperature = 1.00 k + 273.16 = 274.16 k = the change + original temperature = 1.00 k + 273.16 = 274.16 k = the change + original temperature = 1.00 k + 273.16 = 274.16 k P 2 = (6.69 = (6.69 = (6.69 = (6.69 kPa kPa kPa kPa / 273.16 k) x 274.16 k / 273.16 k) x 274.16 k / 273.16 k) x 274.16 k / 273.16 k) x 274.16 k = = = = 6.71449… 6.71449… 6.71449… 6.71449… kPa kPa kPa kPa Change in Pressure = 6.71449… Change in Pressure = 6.71449… Change in Pressure = 6.71449… Change in Pressure = 6.71449… - 6.69 6.69 6.69 6.69 = = = = 0.0244911… 0.0244911… 0.0244911… 0.0244911… Answer = A change in 0.0245 Answer = A change in 0.0245 Answer = A change in 0.0245 Answer = A change in 0.0245 kPa kPa kPa kPa, would indicate a change of 1.0 K. , would indicate a change of 1.0 K. , would indicate a change of 1.0 K. , would indicate a change of 1.0 K. b) b) b) b) Perfect gas, therefore can use same equation from part (a); Perfect gas, therefore can use same equation from part (a); Perfect gas, therefore can use same equation from part (a); Perfect gas, therefore can use same equation from part (a); P 2 = (6.69 = (6.69 = (6.69 = (6.69 kPa kPa kPa kPa / 273.16 k) x / 273.16 k) x / 273.16 k) x / 273.16 k) x 373.16 373.16 373.16 373.16 k k k k = = = = 9.139114072… 9.139114072… 9.139114072… 9.139114072… kPa kPa kPa kPa Answer = 9.14 Answer = 9.14 Answer = 9.14 Answer = 9.14 kPa kPa kPa kPa c) c) c) c) Same Same Same Same steps as in part steps as in part steps as in part steps as in part (a) (a) (a) (a) P 2 = (9.14 = (9.14 = (9.14 = (9.14 kPa kPa kPa kPa / 373.16 k) x 374.16 k / 373.16 k) x 374.16 k / 373.16 k) x 374.16 k / 373.16 k) x 374.16 k = = = = 9.1644935… 9.1644935… 9.1644935… 9.1644935… kPa kPa kPa kPa Change Change Change Change in pressure = 9.164493… in pressure = 9.164493… in pressure = 9.164493… in pressure = 9.164493… - 9.14 = 9.14 = 9.14 = 9.14 = 0.024493… 0.024493… 0.024493… 0.024493… kPa kPa kPa kPa Answer = 0.0245 Answer = 0.0245 Answer = 0.0245 Answer = 0.0245 kPa kPa kPa kPa 2.4(A) 2.4(A) 2.4(A) 2.4(A) A sample consisting of 1.00 mol of perfect gas atoms, for which C V,m = 3/2R, initially at p 1 = 1.00 atm and T 1 = 300 K, is heated reversibly to 400 K at constant volume. Calculate the final pressure, ΔU, q, and w. Solution: This This This This is a perfect gas, thus can use P is a perfect gas, thus can use P is a perfect gas, thus can use P is a perfect gas, thus can use P 1 V 1 /T /T /T /T 1 = P = P = P = P 2 V 2 /T /T /T /T 2 Since Since Since Since this is at constant volume; equation this is at constant volume; equation this is at constant volume; equation this is at constant volume; equation simplifies to simplifies to simplifies to simplifies to: P : P : P : P 1 /T /T /T /T 1 = P = P = P = P 2 /T /T /T /T 2 P 2 = (1.00 / 300) x 400 = = (1.00 / 300) x 400 = = (1.00 / 300) x 400 = = (1.00 / 300) x 400 = 1.333333… 1.333333… 1.333333… 1.333333… atm atm atm atm Answer = Final pressure = 1.33 Answer = Final pressure = 1.33 Answer = Final pressure = 1.33 Answer = Final pressure = 1.33 atm atm atm atm W = 0 because there is no change in volume W = 0 because there is no change in volume W = 0 because there is no change in volume W = 0 because there is no change in volume ΔU = q + w; but w = 0, thus ΔU = q U = q + w; but w = 0, thus ΔU = q U = q + w; but w = 0, thus ΔU = q U = q + w; but w = 0, thus ΔU = q C V,m V,m V,m V,m = 3/2R = (3/2) * 8.31447 = = 3/2R = (3/2) * 8.31447 = = 3/2R = (3/2) * 8.31447 = = 3/2R = (3/2) * 8.31447 = 12.471705 J K 12.471705 J K 12.471705 J K 12.471705 J K -1 mol mol mol mol -1 Cv Cv Cv Cv = = = = 12.471705 J K 12.471705 J K 12.471705 J K 12.471705 J K -1 mol mol mol mol -1 1 1 1 x 1 mol (from the question; 1 mol of perfect) x 1 mol (from the question; 1 mol of perfect) x 1 mol (from the question; 1 mol of perfect) x 1 mol (from the question; 1 mol of perfect) q v = = = = C v ΔT= = = = 12.471705 J K 12.471705 J K 12.471705 J K 12.471705 J K -1 x (400 x (400 x (400 x (400 – 300) = 1247.1705 J 300) = 1247.1705 J 300) = 1247.1705 J 300) = 1247.1705 J Answer = Answer = Answer = Answer = ΔU = ΔU = ΔU = ΔU = q = 1.25 kJ q = 1.25 kJ q = 1.25 kJ q = 1.25 kJ 2.15 2.15 2.15 2.15 Silylene Silylene Silylene Silylene (SiH (SiH (SiH (SiH 2 ) is a key intermediate in the thermal decomposition ) is a key intermediate in the thermal decomposition ) is a key intermediate in the thermal decomposition ) is a key intermediate in the thermal decomposition of of of of silicon hydrides such as silane (SiH 4 ) and disilane (Si 2 H 6 ). Moffat et al. (H.K. Moffat, K.F. Jensen, and R.W. Carr, J. Phys. Chem. 95, 145 (1991)) 95, 145 (1991)) 95, 145 (1991)) 95, 145 (1991)) report report report report Δ f H O (SiH 2 ) = +274 kJ mol -1 . If Δ f H O (SiH 4 ) = +34.3 kJ mol -1 and Δ f H O (Si 2 H 6 ) = +80.3 kJ mol -1 (CRC Handbook (2004)), compute the standard enthalpies of the following reactions: (a) SiH 4(g) →SiH 2(g) + H 2(g) (b) Si 2 H 6(g) →SiH 2(g) + SiH 4(g) Solutions: Solutions: Solutions: Solutions: Use equation below, from the slides (see equation 2.34 on p55 in your Use equation below, from the slides (see equation 2.34 on p55 in your Use equation below, from the slides (see equation 2.34 on p55 in your Use equation below, from the slides (see equation 2.34 on p55 in your text book); text book); text book); text book); (a) (a) (a) (a) [(+274 x 1) [(+274 x 1) [(+274 x 1) [(+274 x 1) + ( + ( + ( + (0 x1 )] 0 x1 )] 0 x1 )] 0 x1 )] – [34.3 x 1] 34.3 x 1] 34.3 x 1] 34.3 x 1] = = = = 239.7 239.7 239.7 239.7 kJ mol kJ mol kJ mol kJ mol -1 -1 -1 -1 (b) (b) (b) (b) [(+274 x 1) + (34.3 x 1)] [(+274 x 1) + (34.3 x 1)] [(+274 x 1) + (34.3 x 1)] [(+274 x 1) + (34.3 x 1)] – [+80.3 x 1] = [+80.3 x 1] = [+80.3 x 1] = [+80.3 x 1] = 228 228 228 228 kJ mol kJ mol kJ mol kJ mol -1 -1 -1 -1