Question

Assume all temperatures to be exact, and neglect significant figures for small changes in dimension. A husband buys a helium-filled anniversary balloon for his wife. The balloon has a volume of 4.8 L in the warm store at 74 ∘F. When he takes it outside, where the temperature is 54 ∘F, he finds it has shrunk

By how much has the volume decreased?

Answer 1

If the mass and the pressure of the gas remains constant, then, from law of ideal gas, we get , (V/T) = constant

where V= volume of the gas and T = temperature of the gas in K

initial volume, V1= 4.8 L

initial temperature, T1 = 74⁰F = (5* ( 74 - 32) / 9 )⁰ C = 23.33⁰ C = (273 + 23.33) K = 296.33 K

final temperature, T2 = 54⁰F= (5* ( 54 - 32) / 9 )⁰ C = 12.22⁰ C = (273 + 12.22) K = 285.22 K

[ Temperature conversion rule used: (i) C/5 = (F-32)/9 & (ii) K = (273 + C)]

Let the final volume be V2

V1 / T1 = V2 / T2

V2 = V1* (T2/T1) = 4.8 * ( 285.22 / 296.33) = 4.62 L

The volume has decresed by ( V1 -V2) = (4.8 - 4.62) L =0.18 L (ans)