A gas occupies 500 cm^3 at normal temperature. At what temperature will the volume of the gas reduced by 20% of its original volume, pressure being constant?
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Answered by
104
Given,
Initial volume of the gas V1 = 500cm3
Normal temperature T1 =273K
Volume is reduced by 20% of the initial volume, i.e volume which mremains is (100-20)% = 80%
Therefore, 80% of 500
=80100×500=400cm3
Thus V2 =400cm3
We have to find T2
From Charles Law we know that,
V1T1= V2T2
Substituting the values we get
500273=400T2
Thus T2 = 273×400500
T2 = 218.4K
Initial volume of the gas V1 = 500cm3
Normal temperature T1 =273K
Volume is reduced by 20% of the initial volume, i.e volume which mremains is (100-20)% = 80%
Therefore, 80% of 500
=80100×500=400cm3
Thus V2 =400cm3
We have to find T2
From Charles Law we know that,
V1T1= V2T2
Substituting the values we get
500273=400T2
Thus T2 = 273×400500
T2 = 218.4K
Answered by
34
Use the formula v1/t1 = v2/t2 hope this help
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