Physics Question ideal gas
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Physics Question ideal gas

[From: ] [author: ] [Date: 12-07-09] [Hit: ]
935=3101.n=0.n=0.R=8.(101325)(1*10^-3)=(0.097)(8.......
An ideal gas has the following initial conditions: Vi = 495 cm3, Pi = 6 atm, and Ti = 100°C. What is its final temperature if the pressure is reduced to 1 atm and the volume expands to 1000 cm3?

What is the equation for this?

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You need to use the Ideal Gas Law: PV=nRT

Pi=6 atm= 607950 N/m^2
Vi=495 cm^3=4.95*10^-4 m^3
Ti=100°C=373 K
R=8.314 J/mol*K

Solve for initial moles:

(607950)(4.95*10^-4)=(8.314)(373)n
300.935=3101.122n
n=0.097 moles

Then uses your new conditions along with the moles to find the new temperature:
Pf=1 atm=101325 N/m^2
Vf=1000 cm^3=1*10^-3 m^3
n=0.097 moles
R=8.314 J/mol*k

(101325)(1*10^-3)=(0.097)(8.314)Tf
101.325=0.806Tf
Tf=125.642K=-147.358°C
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