YF_ISM_19

# 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) IDENTIFY and SET UP: 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 Wp d V = Since p is constant, 2 1 21 () V V Wpd Vp VV == The problem gives T rather than p and V , so use the ideal gas law to rewrite the expression for W . EXECUTE: pV nRT = so 11 1 , pV nRT = 22 2 ; = subtracting the two equations gives pV V nRT T −= Thus Wn R TT =− is an alternative expression for the work in a constant pressure process for an ideal gas. Then ( ) (2.00 mol)(8.3145 J/mol K)(107 C 27 C) 1330 J R = ° ° = + EVALUATE: The gas expands when heated and does positive work. 19.2. IDENTIFY: At constant pressure, . Vn R T =Δ = Δ SET UP: 8.3145 J/mol K. R =⋅ T Δ has the same numerical value in kelvins and in C °. EXECUTE: 3 1.75 10 J 35.1 K. (6 mol) (8.3145 J/mol K) W T nR × Δ= = = KC Δ and 2 27.0 C 35.1 C 62.1 C. T EVALUATE: When 0 W > the gas expands. When p is constant and V increases, T increases. 19.3. IDENTIFY: Example 19.1 shows that for an isothermal process 12 ln( / ). R T pp = pV nRT = says V decreases when p increases and T is constant. SET UP: 358.15 K. T = 3. pp = EXECUTE: (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 ⎛⎞ = ⎜⎟ ⎝⎠ EVALUATE: Since V decreases, W is negative. Figure 19.3 19.4. IDENTIFY: Use the expression for W that is appropriate to this type of process. SET UP: The volume is constant. EXECUTE: (a) The pV diagram is given in Figure 19.4. (b) Since 0, 0. VW = 19

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19-2 Chapter 19 EVALUATE: For any constant volume process the work done is zero. Figure 19.4 19.5. IDENTIFY: Example 19.1 shows that for an isothermal process 12 ln( / ). Wn R T pp = Solve for 1 . p SET UP: For a compression ( V decreases) W is negative, so 518 J. W =− 295.15 K. T = EXECUTE: (a) 1 2 ln . Wp nRT p ⎛⎞ = ⎜⎟ ⎝⎠ / 1 2 . R T p e p = 518 J 0.692. (0.305 mol)(8.314 J/mol K)(295.15 K) W nRT == / 0.692 (1.76 atm) 0.881 atm. R T pp e e = (b) In the process the pressure increases and the volume decreases. The pV -diagram is sketched in Figure 19.5. EVALUATE: 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) IDENTIFY and SET UP: 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. EXECUTE: , 0, V Δ= so 0 W = 23 p is constant; so 53 3 4 (5.00 10 Pa)(0.120 m 0.200 m ) 4.00 10 J WpV =Δ= × = × ( W is negative since the volume decreases in the process.) 4 tot 1 2 2 3 WW W →→ =+= × EVALUATE: The volume decreases so the total work done is negative. 19.7. IDENTIFY: Calculate W for each step using the appropriate expression for each type of process. SET UP: When p is constant, . V When 0, V 0. W = EXECUTE: (a) 13 1 2 1 32 24 2 1 2 41 ( ), 0, ( ) and 0. W p VVW W pVV W = = The total work done by the system is 13 32 24 41 1 2 2 1 () ( ) , WWWW ppVV +++=− which is the area in the pV plane enclosed by the loop.
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## This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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

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