Question

What Celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 23 ∘C , V1= 1.69×103 L ) to a volume of 3.38×103 L ? Assume no change in pressure or the amount of gas in the balloon.

Answer 1

Answer:

319.15^{o}C[/tex]

Explanation:

When all other variables are constant, we are allowed to use the formula

$\frac{T_{2} }{V_{2} } = \frac{T_{1} }{V_{1} } \\Which can be rewritten as T_{2} = \frac{T_{1} V_{2} }{V_{1} }if you make T2 the subject of the formula. This formula is true only if temperature is in Kelvin not degrees Celsius so T1 must be converted to KelvinNow to calculate T2[tex]T_{2}= \frac{296.15K*3.38.10^{3}L }{1.69.10^{3}L }= 592.3K$ = $319.15^{o} C$