Class 9 Chemistry Gas Law A gas at 110kPa at 30.0°C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80°C and the pressure increases to 440Kpa, what is the new volume?

Given, 

P1 = 110kPa , V1 = 2.00 L , T1 = 273+30 = 303°K.

P2 = 440kPa , V2 =? , T2 =273+80 = 353°K.

Solution:

By gas equation:-

P1.V1/T1 = P2.V2/T2.

or , 110×2.00 /303 = 440×V2/353.

or , V2 = (110×2×353)/(440×303) L.

or , V2 = 353/606.L = 0.58250 L or, 582.50 ml. (ANS)