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1 Homework 3 Answer (Due 02/04/2009 Wednesday) P2.3, P2.7, P2.8, P2.11, P2.16, P2.19, P2.20, P2.21, P2.23, P2.25, P2.26 P2.3) 3.00 moles of an ideal gas are compressed isothermally from 60.0 to 20.0 L using a constant external pressure of 5.00 atm. Calculate q, w, Δ U, and Δ H. The work against a constant pressure: () ( ) J 10 03 . 2 m 10 20 m 10 60 Pa 101325 5 V p w 4 3 3 3 3 external × = × × × × = Δ = 0 U = Δ and 0 H = Δ since 0 T = Δ and PV U H + J 10 03 . 2 w q 4 × = = P2.7) For 1.00 mol of an ideal gas, P external = P = 200.0 × 10 3 Pa. The temperature is changed from 100.0°C to 25.0°C, and C V,m = 3/2 R. Calculate q, w, Δ U, and Δ H. J 935 K 373 - K 98 2 mol K J 314472 . 8 mol .0 1 2 3 T C n U 1 - 1 - m V, = × × × = Δ = Δ ( ) J 559 1 K 373 - K 98 2 mol K J 314472 . 8 mol .0 1 2 5 T R C n T C n H 1 - 1 - m V, m p, = × × × = Δ + = Δ = Δ J H 1559 q P = Δ = J 624 ) 1559 ( ) 935 ( q - U P = = Δ = J J w P2.8) Consider the isothermal expansion of 5.25 mol of an ideal gas at 450. K from an initial pressure of 15.0 bar to a final pressure of 3.50 bar. Describe the process that will result in the greatest amount of work being done by the system with P external 3.50 bar and calculate w. Describe the process that will result in the least amount of work being done by the system with P external 3.50 bar and calculate w.

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2 What is the least amount of work done without restrictions on the external pressure? The most work is performed in a reversible process: () J 10 6 . 28 bar 3.50 bar 15.0 ln K 450 mol K J 314472 . 8 mol 5.25 p p ln T R n V V ln T R n w 3 1 - 1 - i f i f reversible × = × × × = = = = f i V V PdV The least amount of work is done in a single stage expansion at constant pressure with the external pressure equal to the final pressure: J bar K mol JK mol nRTP V V P external i f external 3 1 1 i f 10 06 . 15 15.0bar 1 3.50bar 1 ) 50 . 3 ( ) 450 ( ) 314472 . 8 ( ) 25 . 5 ( P 1 P 1 ) ( w × = × × × × = = = The least amount of work done without restrictions on the pressure is zero, which occurs when 0 p external = . P2.11) Calculate Δ H and Δ U for the transformation of 1.00 mol of an ideal gas from 27.0°C and 1.00 atm to 327°C and 17.0 atm if C P , m = 20.9 + 0.042 T K in units of J K 1 mol 1 For an ideal gas, Δ H is given by: + = = K 600 K 300 T T m p, dT K T 0.042 20.9 n dT C n Δ H f i [] J 10 11.94 J 10 67 5. J 10 6.27 J T 0.021 J K 300 K 600 20.9 Δ H 3 3 3 K 600 K 300 2 × = × + × = + × =
3 () J 10 9.45 K 300 K mol J 8.314472 mol 1 J 10 94 . 1 1 Δ T R n Δ H pV Δ Δ H Δ U 3 1 - 1 - 3 × = × × × = = = P2.16) An automobile tire contains air at 320. × 10 3 Pa at 20.0°C. The stem valve is removed and the air is allowed to expand adiabatically against the constant external pressure of 100. × 10 3 Pa until P = P external . For air, C V,m = 5/2 R. Calculate the final temperature. Assume ideal gas behavior.

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