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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?

Respuesta :

dsdrajlin

Answer:

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

asuwafoh

Answer:

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

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