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 of argon (Ar) drops from 50.0 oC to 10.0 oC. Argon is monatomic and may be treated as an ideal gas. (a) What is the change in internal energy of the gas?    3 3 0.450 8.31 10.0 50.0 224 J2 2U nR T       (b) How much work is done by the gas? U = q+w w = -224J Work done by gas, wg = -w = -224 J 10(a) Using pV = nRT n = 30031.8 )101(101.1 35 x xx RT pV   = 0.0441 mol Using pV = nRT, V2 = 5101.1 )375)(31.8(0441.0 xp nRT  = 1.25 x 10-3 m3 10(b) Using pV= nRT, p2 = 3100.1 37531.80441.0  x xx V nRT = 1.37 x 105 Pa

Topic 9: First Law of Thermodynamics Page 3 9749 H2 Physics Tutorial Solutions 11. (i) B to C: constant pressure:  B B T V C C T V  4.1 6606x V TVT B BC C 2830 K D to A: constant volume:  D D T P A A T P  0.1 3008.7 x P TPT A AD D 2340 K (ii) (a) Work done on the gas is given by the force exerted by an exerted agent multiply the displacement. When the gas expands, the displacement is opposite in direction to the force exerted by the external agent thereby giving rise to a negative value of work done on the system. (b) 12 N08/III/4(b) An ideal gas undergoes a cycle of changes A  B C A, as shown in Figure 4.1. Figure 4.1 (i) Calculate the work done by the gas during the change C A. [2]          5 61 10 20 5 10 1.5 Jgw p V Section of cycle Heat supplied to gas / J Work done on gas/J Increase in internal energy of gas/J A to B 0 300 300 B to C 2580 -740 1840 C to D 0 -440 -440 D to A -1700 0 -1700

Topic 9: First Law of Thermodynamics Page 4 9749 H2 Physics Tutorial Solutions (ii) Figure 4.2 is a table of energy changes during one cycle. Complete figure 4.2. Figure 4.2 13. Increase in Internal energy/ J Heat supplied to gas/ J Work done on gas/ J A to B 1200 0 1200 B to C - 1350 -1350 0 C to D - 600 0 -600 D to A 750 750 0 14. In the process illustrated by the pV diagram in Figure 2, the temperature of the ideal gas remains constant at 85.0 oC. Figure 2 (a) How many moles of gas are involved? [0.681 mol] Choose point b:       50.200 1.013 10 0.100 8.31 85 273.15 0.681 mol pV nRT n n         (b) What volume does this gas occupy at a? [0.033 m3]   30.200 0.100 0.033 m0.600 a a b b aPV P V V         (c) How much work was done by or on the gas from a to b? [2250 J] -1.5 4.2 0 -8.5 4.3 5.8

Topic 9: First Law of Thermodynamics Page 5 9749 H2 Physics Tutorial Solutions Isothermal expansion:     0.100ln 0.681 8.31 273.15 85.0 ln 2250 J 0.033 f i Vw nRT V                2250 J of work is done by the gas. (d) By how much did the internal energy of the gas change during this process? [0 J] 3 0 02U nR T U T         15. The graph in Figure 3 shows a pV diagram for 3.25 moles of ideal helium (He) gas. Part ca of this process is isothermal. Figure 3 (a) Find the pressure of the Heat point a. 5 50.040 2.0 10 8.0 10 Pa0.010 a a c c aPV PV P       (b) Find the temperature of the He at points a, b and c. 52.0 10 0.040 296 K8.31 3.25 c c c PVT nR      58.0 10 0.040 1185 K8.31 3.25 b b b P VT nR      296 Ka cT T  (c) How much heat entered or left the He during segments ab, bc and ca? In each did heat enter or leave? ab (isobaric):                          58.0 10 0.040 0.010 24000 J 3 3.25 8.31 1185 296 240002 3 3.25 8.31 1185 296 24000 60000 J2 g ab U q w q q w bc (isochoric):

Topic 9: First Law of Thermodynamics Page 6 9749 H2 Physics Tutorial Solutions    3 3 3.25 8.31 296 1185 36000 J2 2q U nR T        ca (isothermal):     0.010ln 3.25 8.31 296 ln 11100 J 0.040 f i Vq w nRT V               (d) By how much did the internal energy of the He change from a to b, from b to c and from c to a? Indicate whether this energy has increased or decreased.    3 3.25 8.31 1185 296 36000 J2 abU       3 3.25 8.31 296 1185 36000 J2 bcU     0caU  R1 N05/III/02 (a) The ideal gas equation is pV = nRT. Explain why non -SI units may be used for p and V but the temperature cannot have the unit oC. [2] Non-SI units may be used for p and V because we just need a corresponding change in the value for R. R has many possible values depending on the unit system. 273.15K CT T  cannot be corrected in the equation by a s