2.00-mol of a monatomic ideal gas goes from State A to State D via the path A ri
ID: 536417 • Letter: 2
Question
2.00-mol of a monatomic ideal gas goes from State A to State D via the path A rightarrow B rightarrow C rightarrow D: State A P_A = 12.5 atm, V_A = 10.50L State B P_B = 12.5 atm, V_B = 4.50L State C P_C = 21.0 atm, V_C = 4.50L State D P_D = 21.0 atm, V_D = 21.50L Assume that the external pressure is constant during each step and equals the final pressure of the gas for that step. Calculate q for this process. q = Calculate w for this process. w = Use w = -P delta V and sum over the individual steps. Check that you have the correct energy units and be sure you have the sign right (w is defined as the work done ON THE SYSTEM). Calculate delta E for this process delta E = Calculate delta H for this process. delta H =Explanation / Answer
Since nothing is mentioned about the temperature, we will assume that the temperature of the gas stays constant throughout the change.
The pressure-volume work for an ideal gas is given as
w = -PV where P is the constant external pressure and V = Vf – Vi is the change in volume of the gas.
For the process A--->B, we have P = 12.5 atm, V = VB – VA = (4.50 L) – (10.50 L) = -6.00 L.
wAB = -(12.5 atm)*(-6.00 L) = 75 L-atm = (75 L-atm)*(101.325 J/1 L-atm) = 7599.375 J (since 1 L-atm = 101.325 J) = (7599.375 J)*(1 kJ/1000 J) = 7.599375 kJ 7.60 kJ
For the process B-->C, we have P = 21.0 atm, V = VC – VB = (4.50 L) – (4.50 L) = 0.00 L.
wBC = -(21.0 atm)*(0.00 L) = 0.00 L-atm = (0.00 L-atm)*(101.325 J/1 L-atm) = 0.00 J = 0.00 kJ
For the process C-->D, we have P = 21.0 atm, V = VD – VC = (21.50 L) – (4.50 L) = 17.00 L.
wCD = -(21.0 atm)*(17.00 L) = -357.00 L-atm = (-357.00 L-atm)*(101.325 J/1 L-atm)*(1 kJ/1000 J) = -36.173025 kJ -36.17 kJ.
The total work done is w = wAB + wBC + wCD = (7.60 kJ) + (0.00 kJ) + (-36.71 kJ) = -28.57 kJ (ans).
The ideal gas is at a constant temperature. The change in internal energy of an ideal gas, E is only dependent on the temperature change. Since the temperature is constant, T = 0 and hence E = 0 (ans).
We know that
E = q – w where q denotes the heat change.
Since E = 0 and w = -28.57 kJ, therefore,
0 = q – (-28.57 kJ)
===> 0 = q + 28.57 kJ
===> q = -28.57 kJ (ans)
The enthalpy change for an ideal gas depends only on T. As T = 0, hence H = 0 (ans).
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