A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room
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According to charles law ,
V1/T1 = V2/T2
3.5/273+25 = V2/273+40
3.5/298 = V2/313
V2 = 3.5*313/298
V2 = 3.676 L
= 3.68 L approx.
Thus, the new volume of balloon in hot room is 3.68L
Hope it helps.
V1/T1 = V2/T2
3.5/273+25 = V2/273+40
3.5/298 = V2/313
V2 = 3.5*313/298
V2 = 3.676 L
= 3.68 L approx.
Thus, the new volume of balloon in hot room is 3.68L
Hope it helps.
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