Calculate delta S_sys, delta S_surr, and delta S_univ, if 2.00 mol of argon expa
ID: 1059199 • Letter: C
Question
Calculate delta S_sys, delta S_surr, and delta S_univ, if 2.00 mol of argon expands isothermally and reversibly at 298.15 K from a volume of 10.0 L to a volume of 40.0 L. Repeat the calculation when 2.00 mol of argon expands isothermally at 298.15 K from a volume of 10.0 L to a volume of 40.0 L against a constant external pressure of 1.00 atm. delta S_sys (reversible) = delta S_sys (const. ex. P) = delta S_surr (reversible) = delta S_surr (const. ex. P) = delta S_univ (reversible) = delta S_univ (const. ex. P) =Explanation / Answer
Part A: Reversible isothermal process
Consider the isothermal reversible expansion of the gas. The temperature of the gas remains constant. Assuming ideal behavior,
dU = 0 (since the internal energy of an ideal gas is a function of temperature only).
Therefore,
dQ = dU + PdV = PdV
dS = dQ/T
S = 12 dQ/T where 1 and 2 are the initial and final states = 12PdV/T = 12 1/T.(n*R*T/V)(dV) (we use the ideal gas law, PV = n*R*T)
===> S = 12 (n*R).dV/V = n*R*ln(V2/V1)
Plug in V2 = 40.0 L, V1 = 10.0 L, n = 2.0 mol and R = 0.082 L-atm/mol.K and obtain,
Ssys = (2.0 mol)*(0.082 L-atm/mol.K)*ln (40.0/10.0) = 0.2273 L-atm/K= (0.2273 L-atm)*(1 J/0.009869 L-atm) (since 1 J = 0.009869 L-atm) = 23.03 J/K
Since the process is reversible, Ssurr = -Ssys = -23.03 J/K and Suniv = Ssys + Ssurr = (23.03 J/K) + (-23.03 J/K) = 0 (ans).
Part B: Constant pressure process
Ssys = 23.03 J/K (entropy of a system is a state function and depends only on the two states).
Work done by the system, W = -P.V = -(1.00 atm)*(40.0 – 10.0) L = -30.0 L-atm = --(30.0 L-atm)*(1 J/0.009869 L-atm) = -3039.822 J
Q = U – W = -(-3039.822 J) = 3039.822 J
Ssurr = dQ/T = (3039.822/298.15 K) = 10.195 J/K 10.2 J/K
Suniv = Ssys + Ssurr = (23.03 + 10.2) J/K = 33.23 J/K (ans).
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