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Chemistry 12 Online
OpenStudy (anonymous):

n ideal gas in a sealed container has an initial volume of 2.55 L. At constant pressure, it is cooled to 19.00 °C where its final volume is 1.75 L. What was the initial temperature?

OpenStudy (photon336):

constant pressure so re-write the equation. \[\frac{ V_{1} }{ T_{1} } = \frac{ V_{2} }{ t_{2} }\] we're looking for T1 so re-arrange for T1 \[\frac{ V_{1}t_{2} }{ V_{2} } = V_{2}\]

OpenStudy (photon336):

\[\frac{ V_{1}t_{2} }{ V_{2} } = t_{1}\] of course the volume and temperature are proportional to one another. so when one goes up so does the other. \[\frac{ 2.55L*292K }{ 1.75L } = t_{1} = 425Kelvins\] we need to analyze this answer to ensure that it makes sense. The intital volume was higher right? so that means that the initial temperature must be greater. why? because remember you've cooled you system meaning that the T_2 < T_1 and temperature and volume are proportional to one another so the V_2 < V_1

OpenStudy (photon336):

@shelby.lane

OpenStudy (anonymous):

what is it in C

OpenStudy (photon336):

Subtract 273 from the answer

OpenStudy (photon336):

K = 273+C so K-273 = C

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