Thermodynamics A rigid wellinsulated container contains 2.5 kg of Argon at 500 K

Question

Thermodynamics

A rigid wellinsulated container contains 2.5 kg of Argon at 500 K and 4 bar and 1.5kg of Helium at 350 K and 3 bar which are separated by a partition and confined to opposite sides of the container. The partition can freely move and allows conduction from one gas to the other without energy storage in the partition itself. The Argon and Helium each behave as ideal gases with constant specific heat ratio, k=1.4. Determine at equilibrium: (a) The temperature in K. (b)The pressure in bar. (c) The volume occupied by each gas in m^3. Mw Argon = 39.94 kg/kmol Mw Helium = 4.003 kg/kmol

Please indicated steps clearly

Explanation / Answer

Moles of gas, n = mass/ Molecular weight

Moles : Helium = 1.5/4.0003=0.3747kg moles   Argon= 2.5/39.94=0.0626 kg moles

Volume occupied= nRT/P, P is in atm, R= 0.0821 L.atm/mole.K and T= Temperature in K

Volume of helium=0.3747*1000*0.0821*350/(3*0.987)=3636.273 L=3.636m3

Volume of Argon =0.0626*1000*0.0821*500/(4*0.987)=650.89L= 0.65089 m3

Since the container is insulated Q=0 and from first law of thermodynamics

delU= Q+W

delU= W

since there is no work done( the partition can move freely)

delU= 0

let the final temperature be T

moles of helium*(T-350)*CV= moles of Argon*(500-T)*CV

CV= specific heat at constant volume

0.3747*(T-350)= 0.0626*(500-T)

T*(0.3747+0.0626)= 0.0626*500+0.3747*350

T= 371.47 K

From PV= nRT

Where n= 0.3747+0.0626 =0.4373kg moles and V= 3636.273+650.89 L=4287.163 L

R= 0.0821 T= 371.47

P= nRT/V= 0.4373*1000*0.0821*371.47/4287.163=3.11 atm

1 bar= 0.987 atm

3.11 atm= 3.11/0.987=3.151 bar

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