Heat and Thermodynamics. February. 2, Solution of Recitation 2. Consider the first case when air is allowed to expand isothermally.

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1 Heat and Thermodynamics. February., 0 Solution of Recitation Answer : We have given that, Initial volume of air = = 0.4 m 3 Initial pressure of air = P = 04 kpa = Pa Final pressure of air = P = 0 kpa = Pa Consider the first case when air is allowed to expand isothermally. Finial volume < > Initial volume Work done at constant temperature for an ideal gas is given by, According to ideal gas equation and Boyle s law we know that, W = nrt ln P = nrt and P = P = P P Therefore equation can be written as, Application of given data yields, W = P ln P P W = Pa 0.4 m 3 ln Pa Pa = J Now consider the second case in which gas is cooled at constant pressure until it reaches its initial volume. Initial volume < > Finial volume

2 Heat and Thermodynamics. February., 0 Work done at constant pressure is given by, W = P = P = P P = P P P Substitute given values, we get, W = 0.4 m Pa Pa = J Work done in the process of expansion and then cooling will be the sum of these two works. W = J J = J = KJ Total work done is negative, this shows that work done by the system while expanding is greater than the work done on the system while cooling. Answer : We have given that, Initial pressure of gas = P = kpa = 0 3 Pa Initial volume of gas = = 0.7 m 3 Initial temperature of gas = T = 3 0 C = 50 K Final pressure of gas = P = 450 kpa = Pa Finial volume of gas = =.36 m 3 a. Calculations For γ: For an adiabatic process, P γ = Constant P! γ = P γ γ P = P Taking log to both sides, ln ln P P P P = ln = γ ln γ γ = ln P P ln

3 Heat and Thermodynamics. February., 0 Substitute given values we get, γ = ln 03 Pa Pa ln =..36 m m 3 b. Calculations For final temperature T : According to gas laws, Apply given values, P T = P T T = T P P T = 50 K Pa.36 m Pa 0.7 m 3 = 378 K = 04 0 C c. Calculations For number of moles: According to ideal gas equation we know that, P = nrt n = P RT We can calculate number of moles of gas either by using all initial or final values. Consider initial values. where R = 8.34 J/mol.K. n = P RT, n = 03 Pa 0.7 m J/mol.K 50 K = 68 moles d. Calculations For translational kinetic energy: Total translational kinetic energy of the gas before compression is, U initial = 3 nrt = 3 68 moles 8.34 J/mol.K 50 K = J =.96 MJ After compression translational kinetic energy will be, U final = 3 nrt = 3 68 moles 8.34 J/mol.K 378 K = J =.96 MJ e. Calculations For rms speed ratio: Kinetic Energy = 3 kt = mv 3

4 Heat and Thermodynamics. February., 0 Before and after compression, we can write, 3 KT = mv, By dividing above two equations, we get, v v = T T Answer 3: We have given that, v v = 3 kt = mv T 50 K = T 378 K = 0.83 Initial temperature of helium gas = T = 0 0 C = 93 K Initial pressure of helium gas = P = atm. = Pa Initial volume of helium gas = = 8.50 m 3 Final temperature of helium gas = T = 55 0 C = 38 K Final pressure of helium gas = P = P We want to calculate the heat used to raise the temperature of helium gas from initial to final value at constant pressure. for this purpose we shall make use of the following equation. Q = nc P T, where C P = 5 R and number of moles can be calculated by using ideal gas equation. P = nrt n = P RT Consider initial values, Heat absorbed will be, n = P RT Q = P RT 5 R T T = 5P T T T = Pa 8.50 m 3 38 K 93 K 93 K Answer 4: We have given that, = 56 KJ Initial temperature of gas = T Let initial pressure = P Initial volume = Final pressure = P = P Final volume = = 4

5 Heat and Thermodynamics. February., 0 Since pressure and volume are slowly doubled in such a manner as to trace out a straight line on the P diagram, as shown in the figure. P P=P P = Slope of this straight line can be calculated by the ratio of pressure and volume at any two points. Since slope of a straight line is always constant, therefore, P = P P = P a. Work done can be calculated by the equation, f f P W = P d = d = P f d = P i i i = = P f i = P = P = 3 P = 3 nrt b. Internal energy of a gas is given by, U = 3 nrt We want to calculate change in internal energy. U = 3 nr T = 3 nrt T Firstly we shall calculate T. According to ideal gas equation, P = nrt T = P nr Now initial and final temperature of the gas will be, T = P nr and T = P nr = P nr = 4 P nr = 4T f i 5

6 Heat and Thermodynamics. February., 0 Thus internal energy of the gas will be, c From first law of thermodynamics, U = 3 nr4t T = 9 nrt du = dq + dw du = dq + U = Q + W dw Q = U W = 9 nrt + 3 nrt = 6nRT d. Specific heat can be calculated as, C = Q n T = Q nt T = 6nRT n4t T = R Answer 5: From a b: From the figure it is clear that a and b are at the same temperature T. While going from a to b volume decreases and pressure increases, we can say that path ab represents isothermal compression. Work done for isothermal compression is given by, But according to ideal gas equation, W = W ab = f i b P d a P d P = nrt P = nrt 6

7 Heat and Thermodynamics. February., 0 Therefore work will become, W ab = b a nrt d = nrt b a = nrt ln b ln a = nrt ln d = nrt b a ln b a = nrt ln a b From b c: From the figure it is clear that point b and c are at different temperatures, while moving from b to c volume remains constant. Therefore isochoric process is going on between b and c. Since there is no volume change, no work will be done i.e. W bc = From c d: c b P d = P 0 = 0 From the figure it is clear that point c and d are at different temperatures. Temperature of point c is greater than temperature of point d. Thus while moving from c to d temperature decreases. Also here is a decrease in volume, while pressure is constant. Therefore work done at constant pressure will be, W cd = d b P d = P d b = P d b = P d + P b = nrt 3 + nrt From d e: From the figure it is clear that points d and e are at the same temperature, but volume is increasing while pressure is decreasing, therefore path de represents isothermal expansion. Work done for isothermal expansion is, W de = e d = nrt 3 e P d = e nrt 3 d d d = nrt 3 ln d P = nrt e d = nrt 3 ln d e From e a: While going from e to a volume is constant, work done will be zero. a W ea = P d = P 0 = 0 e Total work done along closed path abcdea: 7

8 Heat and Thermodynamics. February., 0 Work done along closed path abcdea will be equal to the sum of all works calculated above. W Total = W ab + W bc + W cd + W de + W ea b = nrt ln + 0 nrt 3 + nrt + nrt 3 ln { a } { }] a d = nr [T + ln + T 3 ln b e d e + 0 For closed reversible cycle, change in internal energy is always equal to zero, therefore, U = Q + W Total = 0 Q = W Total So heat supplied will be equal to total work done. 8