Asked by dave
calculat the temperture of gass when its expanded to 5.25l the gas orignally occupied 3.45l of space at 282k
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Use Charles's law (V1/T1 = V2/T2) assuming pressure and amount of gas are constant.
T2 = T1 * (V2 / V1) = 282 K * (5.25 L / 3.45 L) = 282 * 1.52174 ≈ 429 K.
(≈ 429.1 K; which is about 156 °C.)
T2 = T1 * (V2 / V1) = 282 K * (5.25 L / 3.45 L) = 282 * 1.52174 ≈ 429 K.
(≈ 429.1 K; which is about 156 °C.)
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