00 Bulb Bulb
00 Bulb Bulb
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![]() ProjectionDesign projector bulbs 2 used 400 400 00 400 500 00 US $50.00
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![]() INFOCUS 232 0145 00 232014500 BULB ONLY US $77.32
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![]() Projection Design 400 0400 00 400 0401 00 400 500 00 projector bulb only US $143.50
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![]() New Bulb Philips LCA 3105 00450046004700 Lamp Module US $259.99
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![]() Projector Lamp IPX 400 0401 00 Osram Projector Bulb NEW US $200.00
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![]() 400 0400 00 Bulb ProjectionDesign LampBulb US $156.00
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![]() Original Bare Projector Bulb Part No 400 0402 00 US $95.00
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![]() 151 1039 00 Bulb Runco Projector LampBulb US $156.00
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![]() 313 400 0184 00 3D Perception BULB LAMP US $135.00
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![]() Original 3D Perception 400 0184 00 Projector BulbLamp US $135.00
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![]() Projector Bulb part number 400 0402 00 US $179.00
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![]() 400 0402 00 Projection Design original bulb US $179.00
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![]() Original Projector Bulb 400 0402 00 ProjectionDesign US $179.00
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![]() Original Projector Bulb Part Number 400 0402 00 US $179.00
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![]() 400 0400 00 PROJECTOR ProjectionDesign LampBulb US $156.00
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gas bulb systems?
There are two bulbs attached in a system, where the pressure of the gas in the 1.00 L bulb is 672 mm Hg and that of the gas is the 0.500 L bulb is 442 mm Hg
What is the pressure (atm) of the system when both valves are opened, assuming the temperature remains constant ?
(neglect the volume of the capillary tube connecting the bulbs)
This may not be the way your teacher taught it, but trust me it works.
You can use the ideal gas law for each bulb:
PV = nRT
With a constant T, you can cancel out RT and just solve PV = k, where k is a function of n, the number of moles.
P1 * V1 = k1 and P2 * V2 = k2
k1 = 672 mm Hg * 1.00 L = 672. mm Hg * L
k2 = 442 mm Hg * 0.500 L = 221 mm Hg * L
You can add these together: k1 + k2 = k3 = 893 mm Hg * L
The final volume is 1.5 L
P3 = k3/V3 = 893 mm Hg * L / 1.5 L = 595.3 mm Hg
convert that to atmospheres, multiply by 0.001316 atm/mm Hg = 0.783 atm


US $50.00































