At a constant pressure and a temperature of 280 K, the gas in a cylinder has a volume of 10.0 liters. If the volume of the gas is increased to 20.0 Liters what is the new temperature?
V1/T1=V2/T2 so T1/V1V2= 280K/10l20l=560K. Weird question, but this should be the answer
Answer:
560K
Explanation:
For the question above we apply Charles law.
Charles law states that at constant pressure, the Volume of an ideal gas is directly proportional to it's temperature in Kelvin(Absolute Temperature).
Charles law is represented mathematically by:
V₁/ T₁ = V₂/T₂
Where V₁= 10.0 liters , T₁ = 280K
V₂= 20.0 litres, T₂ = ?
Therefore, 10/280 = 20/T₂
Cross multiplying
10× T₂= 280×20
T₂= (280×20)÷ 10
T₂ = 560K
The new temperature is 560K.