Calculate the change in entropies of the system and the surroundings, and the to

Question

Calculate the change in entropies of the system and the surroundings, and the total change in entropy, when a sample of nitrogen gas of mass 35 g at 298 K and 1.00 bar doubles its volume in the following situations. an isothermal reversible expansion Delta S(gas) J K-1 Delta S(surroundings) J K-1 Delta S( total) J K-1 an isothermal irreversible expansion against pex = 0 Delta S(gas) J K-1 Delta S(surroundings) J K-1 Delta S( total) J K-1 an adiabatic reversible expansion Delta S(gas) J K-1 Delta S(surroundings) J K-1 Delta S( total) J K-1

Explanation / Answer

I will assume that nitrogen gas behaves as an ideal gas. For the question (a), remember that the internal energy of an ideal gas depends only on temperature, so in an isothermal process, the internal energy of an ideal does not change. From the first law, we know that: dE = ?q - ?w, where ?q is the thermal energy added to the system and ?w is the work done by the system on the surroundings. Here, dE = 0 for an isothermal process involving an ideal gas, so: ?q = ?w For strictly mechanical work, ?w = pdV, and for an ideal gas p = n*R*T/V, so: ?q = n*R*T/V dV The second law tells us that in general: dS ? ?q/T, and in the special case of a reversible process: dS = ?q/T. So, the entropy change of the nitrogen gas is given by: dS = n*R/V dV Integrating this yields: ?S = n*R*ln(V_final/V_initial) The atomic mass of N is 14 gm/mol, so 35 gm N = 2.5 mol. The entropy change of the sample would be: ?S_sample = (2.5 mol)*(8.314 J/(mol*K))*ln(2) (as volumes doubles ) ?S_sample = +14.407 J/K Now, the thermal energy added to the N sample was provided by the surroundings, so dS_surr = -?q/T = -dS_sample, and ?S_surr = -?S_sample = -14.407 J/K. The entropy change of the universe is the sum of ?S_surr and ?S_sample: ?S_total = ?S_sample + ?S_surr = 0 The entropy change of the universe is zero for reversible processes. -------------- For question (b), note that the N sample ends up in exactly the same state (temperature and volume) as in part (a). Entropy is a state function, (i.e., the entropy of a system at equilibrium depends only on the state of the system, not on the history of the system. The entropy difference between two states does not depend on the path taken to get from one state to another), so the entropy change of the N sample must be the same as in part (a). ?S_sample = +14.407 J/K This is again specified to be an isothermal expansion, so again the change in internal energy is zero. 0 = ?q - ?w However, for a free expansion, p_ext = 0, so ?w = pdV = 0: 0 = ?q No thermal energy is exchanged between the system and the surroundings (i.e., this is also an adiabatic process), so the entropy change of the surroundings is given by: dS ? ?q/T = 0 ?S_surr ? 0 Note that we have to use the inequality in this case because the expansion was an irreversible process. The entropy change of the universe in for the irreversible free expansion is therefore: ?S_total ? +14.407 J/K -------------------------------- For question (c), we are dealing with an adiabatic rather than an isothermal process. In an adiabatic process, ?q = 0, and if the process is also reversible, (as it is in this case), this means that dS = ?q/T = 0; the entropy changes of both the surroundings and the sample are zero, and so is the entropy change of the universe (?S_total is always zero for reversible processes). Please Give Chegg points.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.