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?

1 year ago
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 \[\frac{ V_{1}t_{2} }{ V_{2} } = V_{2}\]

1 year ago
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

1 year ago
OpenStudy (photon336):

@shelby.lane

1 year ago
OpenStudy (anonymous):

what is it in C

1 year ago
OpenStudy (photon336):

Subtract 273 from the answer

1 year ago
OpenStudy (photon336):

K = 273+C so K-273 = C

1 year ago