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YF_ISM_19 - THE FIRST LAW OF THERMODYNAMICS 19 The...

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19-1 T HE F IRST L AW OF T HERMODYNAMICS 19.1. (a) I DENTIFY and S ET U P : The pressure is constant and the volume increases. The pV -diagram is sketched in Figure 19.1 Figure 19.1 (b) 2 1 V V W p dV = Since p is constant, 2 1 2 1 ( ) V V W p dV p V V = = The problem gives T rather than p and V , so use the ideal gas law to rewrite the expression for W . E XECUTE : pV nRT = so 1 1 1 , pV nRT = 2 2 2 ; p V nRT = subtracting the two equations gives 2 1 2 1 ( ) ( ) p V V nR T T = Thus 2 1 ( ) W nR T T = is an alternative expression for the work in a constant pressure process for an ideal gas. Then 2 1 ( ) (2.00 mol)(8.3145 J/mol K)(107 C 27 C) 1330 J W nR T T = = ° ° = + E VALUATE : The gas expands when heated and does positive work. 19.2. I DENTIFY : At constant pressure, . W p V nR T = Δ = Δ S ET U P : 8.3145 J/mol K. R = T Δ has the same numerical value in kelvins and in C °. E XECUTE : 3 1.75 10 J 35.1 K. (6 mol) (8.3145 J/mol K) W T nR × Δ = = = K C T T Δ = Δ and 2 27.0 C 35.1 C 62.1 C. T = ° + ° = ° E VALUATE : When 0 W > the gas expands. When p is constant and V increases, T increases. 19.3. I DENTIFY : Example 19.1 shows that for an isothermal process 1 2 ln( / ). W nRT p p = pV nRT = says V decreases when p increases and T is constant. S ET U P : 358.15 K. T = 2 1 3 . p p = E XECUTE : (a) The pV -diagram is sketched in Figure 19.3. (b) 1 1 (2.00 mol)(8.314 J/mol K)(358.15 K)ln 6540 J. 3 p W p = = − E VALUATE : Since V decreases, W is negative. Figure 19.3 19.4. I DENTIFY : Use the expression for W that is appropriate to this type of process. S ET U P : The volume is constant. E XECUTE : (a) The pV diagram is given in Figure 19.4. (b) Since 0, 0. V W Δ = = 19
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19-2 Chapter 19 E VALUATE : For any constant volume process the work done is zero. Figure 19.4 19.5. I DENTIFY : Example 19.1 shows that for an isothermal process 1 2 ln( / ). W nRT p p = Solve for 1 . p S ET U P : For a compression ( V decreases) W is negative, so 518 J. W = − 295.15 K. T = E XECUTE : (a) 1 2 ln . W p nRT p = / 1 2 . W nRT p e p = 518 J 0.692. (0.305 mol)(8.314 J/mol K)(295.15 K) W nRT = = − / 0.692 1 2 (1.76 atm) 0.881 atm. W nRT p p e e = = = (b) In the process the pressure increases and the volume decreases. The pV -diagram is sketched in Figure 19.5. E VALUATE : W is the work done by the gas, so when the surroundings do work on the gas, W is negative. Figure 19.5 19.6. (a) I DENTIFY and S ET U P : The pV -diagram is sketched in Figure 19.6. Figure 19.6 (b) Calculate W for each process, using the expression for W that applies to the specific type of process. E XECUTE : 1 2, 0, V Δ = so 0 W = 2 3 p is constant; so 5 3 3 4 (5.00 10 Pa)(0.120 m 0.200 m ) 4.00 10 J W p V = Δ = × = − × ( W is negative since the volume decreases in the process.) 4 tot 1 2 2 3 4.00 10 J W W W = + = − × E VALUATE : The volume decreases so the total work done is negative. 19.7. I DENTIFY : Calculate W for each step using the appropriate expression for each type of process. S ET U P : When p is constant, . W p V = Δ When 0, V Δ = 0. W = E XECUTE : (a) 13 1 2 1 32 24 2 1 2 41 ( ), 0, ( ) and 0. W p V V W W p V V W = = = = The total work done by the system is 13 32 24 41 1 2 2 1 ( )( ), W W W W p p V V + + + = which is the area in the pV plane enclosed by the loop.
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