Asked by ed
the volume of a gas was originally 2.5 L; it's pressurewas 104kpa and its temperatire was 270K. the volume of the gas expanded to 5.3 L and its pressure decreased to 95 kpa. What is the temperature of the gas?
Answers
Answered by
DrBob222
(P1V1/T1) (P2V2/T2)
T must be in kelvin.
T must be in kelvin.
Answered by
Jai
For an ideal gas,
(P1*V1)/(T1) = (P2*V2)/(T2)
where
P1, V1 and T1 are the pressure, volume and temp at initial conditions, and
P2, V2 and T2 are the final conditions
Substituting,
(104*2.5)/270 = (95*5.3)/(T2)
now solve for T2.
(P1*V1)/(T1) = (P2*V2)/(T2)
where
P1, V1 and T1 are the pressure, volume and temp at initial conditions, and
P2, V2 and T2 are the final conditions
Substituting,
(104*2.5)/270 = (95*5.3)/(T2)
now solve for T2.
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