(d). If I let thesame gas expand by removing all external pressure, how much wor

Question

(d). If I let thesame gas expand by removing all external pressure, how much work isdone?

Explanation / Answer

a) W = -nRT*ln[V2/V1     isothermal work W = -(0.1 mol * 8.314 J/mol/K * 304 K ) *ln(0.5 / 1 L) =175.1899 Joules ~ 175 J U = 0 b/c T is constant. H = 0 b/c of samereasoning U = q + W = 0   => q = -W = -175 J b) P = (nRT)/V = (0.1 mol * 0.0821 L*atm/mol/K * 304 K) / (0.5L)                       Use V = 0.5 L, b/c then Pgas = Pext P = 4.99168 atm P = 505781.976 Pa = 5.058 x 105 Pa W = -P*V = -5.058e5 Pa * ( 0.5 L - 1 L) *(1 m3 /1000 L) =252.890988 J = 253 Joules same thing about U and H = 0 b/c T= constant Q = -W = -253 J c) Pressure of the gas is the same as the external pressure; the forceon the piston is zero d) Pext = 0 W = integral [-Pext*dV] = 0

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