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A birthday balloon had a volume of 14.1 L when the gas inside was at a temperature of 13.9 °C. Assuming no gas escapes, what is its volume when the balloon warms up to 22.0 °C?

Respuesta :

Eduard22sly

Answer:

14.5L

Explanation:

The following data were obtained from the question:

V1 = 14.1L

T1 = 13.9°C = 13.9 + 273 = 286.9K

T2 = 22°C = 22 + 273 = 295K

V2 =?

Using charles' law: V1/T1 = V2 /T2, we can obtain the new volume as follows:

14.1/286.9 = V2 /295

Cross multiply to express in linear form

286.9 x V2 = 14.1 x 295

Divide both side by 286.9

V2 = (14.1 x 295) / 286.9

V2 = 14.5L

Therefore, the new volume = 14.5L

Cricetus

"14.5 L" would be the volume when the balloon warms up to 22.0°C.

Given:

Volume,

  • [tex]V_1 = 14.1 \ L[/tex]
  • [tex]V_2 = ?[/tex]

Temperature,

  • [tex]T_1 = 13.9^{\circ} C[/tex] or, [tex]286.9 \ K[/tex]
  • [tex]T_2 = 22^{\circ} C[/tex] or, [tex]295 \ K[/tex]

By using Charles' Law, we get

→ [tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

or,

→ [tex]V_2 = \frac{V_1\times T_2}{T_1}[/tex]

By substituting the values, we get

→     [tex]= \frac{14.1\times 295}{286.9}[/tex]

→     [tex]= 14.5 \ L[/tex]

Thus the above approach is appropriate.

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