Consider a sample containing 2.00 mol of a monatomic ideal gas that undergoes th

Question

Consider a sample containing 2.00 mol of a monatomic ideal gas that undergoes the following changes: PA = 10.0 atm PB = 10.0 atm PC = 20.0 atm VA = 10.0 L rightarrow VB = 5.0 L rightarrow VC = 5.0 L PD = 20.0 atm rightarrow VD = 25.0 L For each step, assume that the external pressure is constant and equals the final pressure of the gas for that step. Calculate q, w, Delta E, and Delta H for each step and for the overall change from state A to state D.

Explanation / Answer

PV =nRT            T = PV/nR TA = 10 atm*10L/(2 mol *0.0821 L*atm/mol/K) = 609.01 K TB = 10 atm* 5L/(2mol*0.0821 L*atm/mol/K) = 304.51 K TC = 20 atm*5L/(2mol *0.0821 L*atm/mol/K) = 609.01 K TD = 20 atm*25 L/(2mol*0.0821L*atm/mol/K) = 3045.07 K Equations:    delU = Cv*delT;   delH =Cp*delT;    W = -P_ext*delV   if externalpressure is constant; delU = Q + W for monatomic ideal gas; Cv = 3nR/2;   Cp = 5nR/2 atclassical temperatures A -> B delU = (3*nR/2)*(TB-TA) = (3*2*8.314 J/mo/K /2)*(304.51 - 609.01)K= -7594.839 J delH = (5*nR/2)*(TB-TA) = -12658.065 J W = -10.0 atm*(5L-10L)*(1e-3 m3/1 L) *(101325 Pa/1atm) *(1J/(1Pa*m3)) = 5066.25 J Q = delU - W = -7594.839 J - 5066.25J = -12661.089 J B -> C delU = (3*nR/2)*(TC-TB) = (3*2*8.314 J/mo/K /2)*(609.1 - 304.51)K =7594.839 J delH = (5*nR/2)*(TC-TB) = 12658.065 J W = -20.0 atm*(5L-5L)*(1e-3 m3/1 L) *(101325 Pa/1atm) *(1J/(1Pa*m3)) = 0 J Q = delU - W = 7594.839 J C - > D delU = (3*nR/2)*(TD-TC) = (3*2*8.314 J/mo/K /2)*(3045.07 - 609.01)K= 60760.2085 J delH = (5*nR/2)*(TD-TC) = 101267.014 J W = -20.0 atm*(25L-5L)*(1e-3 m3/1 L) *(101325 Pa/1atm) *(1J/(1Pa*m3)) = -40530 J Q = delU - W = 101290.209 J Overall A -> D delU = -7594.839 J + 7594.839 J + 60760.2085 J = 60760.2085 J delH = -12658.065 J + 12658.065 J + 101267.014 J = 101267.014 J W = 5066.25 J + 0 J + (-40530 J) = -35463.75 J Q = -12661.089 J + 7594.839 J + 101290.209 J = 96223.959 J

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