H209 First Law of Thermodynamics 2.1 Tutorial Solution (1718)

Topic 9: First Law of Thermodynamics Page 1 9749 H2 Physics Tutorial Solutions First Law of Thermodynamics Answer Key 1. D 10. (a) 1.25 x 10-3 m3 R1(b) 273.15 K 2. D 10. (b) 1.37 x 105 Pa R1(c)(i) 1.35x106 m3 3. D 14.(a) 0.681 mol R1(ii) 1.77x105 N 4. E 14.(b) 0.033 m3 R1(iii) 1.77x105 N 5. B 14.(c) 2250 J R1(iv) 3.99x103 N 6. -6 J 14(d) 0 R1(v) 0.221 ms-2 7(a) 3.78 x 104 J 15.(a) 8.0x105 Pa R1(d)(ii) 4.72x10-21J 7(b) 7.72 x 104 J 15.(b) 296 K R1(iii) 6.33x105 mol 8(a) 1.66 x 105 J 15.(c) qab =60,000J R1(iv) 1.80x109 J 8(b) 2.03 x 106 J qbc = -36,000 J qca = -11100 J R2(b)(i) 0.295 mol 9(a) -224 J 15(.d) Uab = 36000 J R2(c)((ii) [3.71 x 103 J] 9(b) 224 J Ubc = 36000 J R2(c)((iii) 1100 J Uca = 0 6. N07/I/20 An ideal gas undergoes an expansion in volume from 1.3 x 10-4 m3 to 3.6 x 10-4 m3 at a constant pressure of 1.3 x 105 Pa. During this expansion, 24 J of heat is supplied to the gas. What is the overall change in the internal energy of the gas? [-6 J] Isobaric expansion:     5 41.3 10 3.6 1.3 10 29.9 Jw p V             First law:  24 29.9 5.9 JU q w        7. A gas in a cylinder expands from a volume of 0.110 m 3 to 0.320 m3. Heat flows into the gas just rapidly enough to keep the pressure constant at 1.80 x 10 5 Pa during the expansion. The total heat added is 1.15 x 105 J. (a) Find the work done by the gas. [3.78 x 104 J]          5 41.80 10 0.320 0.110 3.78 10 Jgw p V (b) Find the change in internal energy of the gas. [7.72 x 104 J]          5 4 41.15 10 ( 3.78 10 ) 7.72 10 JU q w

Topic 9: First Law of Thermodynamics Page 2 9749 H2 Physics Tutorial Solutions 8. When water is boiled under a pressure of 2.00 atm, the heat of vaporization is 2.20 MJ kg-1 and the boiling point is 120 oC. At this pressure and temperature, 1.00 kg of water has a volume of 1.00 x 10-3 m3, and 1.00 kg of steam has a volume of 0.824 m3. [1 atm = 1.013 x 105 Pa] (a) Compute the work done when 1.00 kg of steam is formed at this temperature.           5 52 1.013 10 0.824 0.001 1.67 10 Jgw p V (b) Compute the increase in internal energy of the water.          6 5 62.20 10 ( 1.67 10 ) 2.03 10 JU q w (c) The work done for 1.00 kg of steam to form at 100 oC and 1 atm is 1.69 x 10 5 J, suggest a reason why the work done against external pressure decreases when external pressure increases. The work done has decreased at higher pressure because the volume occupied by the water vapour has decreased by a factor larger than the factor increase in pressure. E.g. At 1 atm; volume occupied by 1 kg of steam is 1.671 m3. 1 atm 2 atm 1.671 2.0280.824 Vol Vol   [1.67 x 105 J, 2.03 x 106 J] 9. During an adiabatic expansion, the temperature of 0.450 mol