Consider a sample containing 2.00 mol of a monoatomic ideal gasthat undergoes th
ID: 684950 • Letter: C
Question
Consider a sample containing 2.00 mol of a monoatomic ideal gasthat undergoesthe following changes:
Step 1: (PA = 10.0 atm, VA = 10.0 L) to (PB = 10.0 atm, VB = 5.0L)
Step 2: (PB = 10.0 atm, VB = 5.0 L) to (PC = 20.0 atm, VC = 5.0L)
Step 3: (PC = 20.0 atm, VC = 5.0 L) to (PD = 20.0 atm, VD = 25.0L)
For each step, assume that the external pressure is constant andequals the final
pressure of the gas for that step. Calculate q, w, E, andH for each step and
for the overall change from state A to state D.
Step 1 Step 2 Step 3 Total q 1 kJ 2 kJ 3 kJ 4 kJ w 5 kJ 6 kJ 7 kJ 8 kJ E 9 kJ 10 kJ 11 kJ 12 kJ H 13 kJ 14 kJ 15 kJ 16 kJ
Step 1 Step 2 Step 3 Total q 1 kJ 2 kJ 3 kJ 4 kJ w 5 kJ 6 kJ 7 kJ 8 kJ E 9 kJ 10 kJ 11 kJ 12 kJ H 13 kJ 14 kJ 15 kJ 16 kJ
Explanation / Answer
T = PV/nR TA = (10 atm*10 L)/(2mol *0.0821 L*atm/mol/K) = 609.013398 K TB = (10 atm*5 L)/(2mol *0.0821 L*atm/mol/K) = 304.506699 K TC = (20 atm*5 L)/(2mol *0.0821 L*atm/mol/K) = 609.013398 K TD = (20 atm*25 L)/(2mol *0.0821 L*atm/mol/K) = 3045.06699 K E =n*Cv*T Cv = (3*R/2) H = E + nRT w = -p_ext*V E = q + w => q = E - w Step 1: (PA = 10.0 atm, VA = 10.0 L) to (PB = 10.0 atm, VB = 5.0L) E = (3nR/2)*T = (3*2mol*8.314 J/mol/K)/2 *(304.506699 K - 609.013398 K) = -7595.00609 J H = E + nRT = -7595.00609 J +2*(8.314)*(304.506699 - 609.013398) = -12658.3435 J w = -10.0 atm*(5 - 10)L *(1 m3/1000L)*(101325 Pa/ 1atm) = 5066.25J q = E - w = -7595.00609 J - 5066.25 J = -12661.2561 J Step 2: (PB = 10.0 atm, VB = 5.0 L) to (PC = 20.0 atm, VC = 5.0L) E = (3nR/2)*T = (3*2mol*8.314 J/mol/K)/2 *(609.013398 K-304.506699 K) = 7595.00609 J H = E + nRT = -7595.00609 J +2*(8.314)*(609.013398 K-304.506699 K) = 12658.3435 J w = -20.0 atm*(5 - 5)L = 0 q = E - w = 7595.00609 J - 0 = 7595.00609 J Step 3: (PC = 20.0 atm, VC = 5.0 L) to (PD = 20.0 atm, VD = 25.0L) E = (3nR/2)*T = (3*2mol*8.314 J/mol/K)/2 *(3045.06699K -609.013398 K) = 60760.0487 J H = E + nRT = -7595.00609 J +2*(8.314)*(3045.06699K -609.013398 K) = 101266.748 J w = -20.0 atm*(25 - 5)L*(1 m3/1000L)*(101325 Pa/1atm) = -40530J q = E - w = 60760.0487 J - ( -40530 J) = 101290.049 J ====================== Step1 Step2 Step3 Total q -1.27e4J 7.60 e3J 1.01e5J 9.60e4 J w 5.07e3J 0J -4.05e4 J -5.56e4 J E -7.60 e3J 7.60 e3J 6.08 e4J 6.08e4 J H -1.27 e4J 1.27 e4J 1.01e5J 1.01e5 J in kJ , recall that 1e3 J = 1kJ Step1 Step2 Step3 Total q -13 kJ 7.6kJ 1.0 e2kJ 96 kJ w 5.1 kJ 0kJ -41 kJ -56 kJ E -7.6kJ 7.6kJ 61 kJ 61 kJ H -13kJ 13kJ 1.0 e2kJ 1.0 e2 kJ
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