A sample of gas in a balloon has an initial temperature of 13 ∘C and a volume of 1.04*103 L. If the temperature changes to 56 ∘C, and there is no change of pressure or amount of gas, what is the new volume, V2V_2, of the gas?

1.196*10³ L

Explanation:

Since the pressure is constant,

We can use Charles's law

V/T = V'/T' ... Equation 1

Where V = Initial Volume, T = Initial Temperature, V' = Final Volume, T' = Final Temperature.

make V' the subject of the equation

V' = T' (V/T) ... Equation 2

Given: V = 1.04*10³ L, T = 13 °C = (273+13) K = 286 K, T' = 56 °C = (273+56) = 329 K

Substitute into equation 2

V' = 329 (1.04*10³) / 286

V' = 1.15 (1.04*10³)

V' = 1.196*10³ L.

Hence the new volume = 1.196*10³ L

1.20 * 10³ L

Explanation:

Given data

Initial temperature (T₁) : 13°C + 273.15 = 286 K Initial volume (V₁) : 1.04 * 10³ L Final temperature (T₂) : 56°C + 273.15 = 329 K Final volume (V₂) : ?

We can find the final volume using Charle's law.

V₁/T₁ = V₂/T₂

V₂ = V₁ * T₂/T₁

V₂ = 1.04 * 10³ L * 329 K/286 K

V₂ = 1.20 * 10³ L