In: Chemistry
Consider a 10.00 L cylinder divided by an adiabatic piston. Side
A contains Ar at PA1 = 2.00
atm, VA1 = 5.00 L, and TA1 = 500K. Side B contains Ne at PB1 = 2.00
atm, VB1 = 5.00 L,
and TB1 = 500K. Side B is completely adiabatic and closed. Side A
is maintained isothermal
and has a small hole in the end, which is plugged by a stopper.
When the stopper is pulled,
the Ar in Side A can be pushed out by the pressure from Side B
until the pressure on Side A
equals the pressure outside (1.00 atm).
(a) Calculate the final conditions on both sides.
(b) Calculate q, w, ∆U, ∆H, and ∆S for both sides. You can assume
that the process is
reversible.
A ) Ideal gas PV = n R T R = universal gas constant = 8.314 J/mole K
side A ; Ne , V1 = 5L = 0.005 m3 , n= 6.22 moles , T = 350 K ( Isothermal) , V2 = 0.5 V1 = 0.0025 m3
for isothemal process P1V1 = P2V2
P1 = ( n R T) / V1 = 3620 KPa
P2 = 3620 * 0.005 / 0.0025 = 7240 Kpa
Side B He ( monoatomic gas , = 1.67 ) V1 = 5 L = 0.005 m3 , n = 4.5 moles V2 = 2 V1 = 0.01
Rapid heating is a adiabatic process , so P1 V1 = P2 V2 = C ( eq 1)
Work done in Side B = Work done for compression in Side A
= P1V1 ln ( V2/ V1) (side A ) = - 12.55 Kw
i.e ( P1V1 - P2V2) / ( -1 ) ( Side B ) = 12 .55 Kw
P1V1 - P2V2 = 8.4 ( P in Kpa) (eq 2)
on solving eq 1 and 2
P2 ( 21.67 * 0.005 - 0.01 ) = 8.4 P2 = 1421.14 KPa
P1 = 4522.27 Kpa
B) Side A ; Entropy change in Isothermal process = dS = dQ / T = nR ln (V2/V1) = -35.844 J /K
Side B Adiabatic process dQ =0 , dS = 0