Gas occupies a volume of 2 m3 at a temperature of 0 degrees Celsius. What will its volume be at a temperature of 27 degrees Celsius and the same pressure?
V1 = 2 m3.
t1 = 0 ° C.
t2 = 27 ° C.
P = const.
For a constant mass of gas, at a constant pressure, an isobaric process occurs, which is described by the Gay-Lussac law: the ratio of the volume of gas V to its absolute temperature T remains constant.
V1 / T1 = V2 / T2.
V2 = V1 * T2 / T1.
We express the absolute temperature by the formula: T = t + T.
T1 = t1 + 273.
T1 = 0 ° C + 273 = 273 ° K.
T2 = t2 + 273.
T2 = 27 ° C + 273 = 300 ° K.
V2 = 2 m3 * 300 ° K / 273 ° K = 2.2 m3.
Answer: the gas volume will increase to V2 = 2.2 m3.
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