Example Problems for Week 3 with Solutions

25 ( ; 8) = 25L
atm Ws = 25 (  + 8) + 2 2 By this method the work done by the system is 25  101:3J = 7956J . The second method involves examining the area under the curve, a circle with radius 5. So the area is 25L
atm. This again is the work done by the system equal to 7956J . The maximum and minimum temperatures can be found from the ideal gas law for one mole of gas, PV = RT . Tmax attains when (PV ) is a maximum and Tmin attains when (PV ) is a minimum. Let P 0 = P ; 10 and V 0 = V ; 10 so PV is an extremal when P 0V 0 is extremal. p P 0V 0 = f (P 0(V 0); V 0) =  25 ; V 02 V 0 df = 0 = p25 ; V 02  V 0 1 p ;2V 0 dV 0 2 25 ; V 02# " 02 02 0 =  25p; V ; 02V 25 ; V q The extrema are for V 0 =  252 . Examination or checking the sign of the second derivative, d2 F d2 F dV 2 > 0 for a minimum and dV 2 < 0 for a maximum, gives the following. 0 [h] qV 0 Vq Pq qP 25 25 25 max q2 10 + q 2 10 + q 252 q2 min ; 252 ; 252 10 ; 252 10 ; 252 This leads to Tmax = (PV )max=R = 2203K and Tmin = (PV )max=R = 502K . If the number of moles are doubled the work done by the system is still W = 7956J but the maximum and minimum temperatures are decreased by a factor of two from T = PV=(nR) and Tmax = 1102K and Tmin = 251K . Example Problem 3.2 One mole of ideal gas at 25C and 1atm undergoes the following reversible process: 1. Isothermal expansion to 0:5atm, followed by 2. Isobaric expansion to 100C, followed by 3. Isothermal compression to 1atm, followed by 4. Isobaric compression to 25C The system then undergoes the following cyclic process: 1. Isobaric expansion to 100C, followed by 2. A decrease in pressure at constant volume to P atm, followed by 3. An isobaric compression at P atm to 24:5L, followed by 4. An increase in pressure at constant volume to 1atm. Calculate the value of P which makes the work done on the gas in the rst cycle equal to the work done by the gas in the second cycle. Solution 3.2 R For one mole of an ideal gas PV = RT and W = PdV . W is the work done by the gas. There are 3 kinds of processes here and the work must be calculated accordingly. 1. For an isothermal process (T is constant): P = RT=V so W = RT ln VV01 = RT ln PP01 . 2. For an isobaric process (P is constant): W = P (V1 ; V0 ) = R(T1 ; T0 ) 3. For a constant volume process (one of the isochoric processes) W = 0 but the pressure and temperature will change. Let's begin by summarizing the transformations of the gas by recording the T,P,V before and after each process in liters, atmospheres and Kelvin and also the form of the work done by the system. T0 298 298 373 373 298 373 T3 . P0 1 0:5 0:5 1 1 1 Pu Pu V0 . . . . . Vu Vu 24:5 T1 298 373 373 298 373 T3 T4 . P1 0:5 0:5 1 1 1 Pu Pu 1 V1 . . . . Vu Vu 24:5 24:5 Wi W1 = RT ln PP01 W2 = R(T1 ; T0 ) W3 = RT ln PP01 W4 = R(T1 ; T0 ) W10 = R(T1 ; T0 ) W20 = 0 0 W3 = R(T1 ; T0 ) W40 = 0 For the rst reversible process the work done by the system is W = P4i=1 Wi. W= 4 X i=1 Wi = 298R ln 2 + R(373 ; 298) + 373R ln 0:5 + R(298 ; 373) W = (298 ; 373)R ln 2 = ;75R ln 2 For the second cyclic process the work done by the system is W 0 = P4i=1 Wi0. W0 = 4 X i=1 Wi0 = R(373 ; 298) + 0 + R(T4 ; T3 ) + 0 W 0 = 75R + R(T4 ; T3 ) So we must nd the unknown temperatures using T = PV=R and V = RT=P . T3 = PRu Vu = PRu 3731 R = 373Pu T4 = PuR24:5 Now W 0 = 75R + R( PuR24:5 ; 373Pu) = 75R + RPu( 24R:5 ; 373). The restriction that the work done on the gas in the rst cycle equals the work done by the gas in the second cycle is equivalent to ;W = W 0. W + W 0 = 0 = ;75R ln 2 + 75R + RPu( 24R:5 ; 373) 2 ; 1) Pu = 75(ln 24:5 ; 373 R J
1atm5
103 L3 = 0:08207 L
atm . So Pu = 0:3atm. In the above R = 8:314 moleK 1:01310 Pa m moleK