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A helium-filled weather balloon has a volume of 793 L at 16.9°C and 759 mmHg. It is released and rises to an altitude of 4.05 km, where the pressure is 537 mmHg and the temperature is –7.1°C.

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nkirotedoreen46

Answer:

1082.96 L

Explanation:

We are given;

  • Initial volume of helium gas, V1 = 793 L
  • Initial temperature, T1 = 16.9°C

            But, K = °C + 273.15

  • Thus, initial temperature, T1 is 290.05 K
  • Initial pressure, P1 = 759 mmHg
  • New pressure at 4.05 km, P2 = 537 mmHg
  • New temperature at 4.05 km, T2 = 7.1 °C

                                                             = 280.25 K

Assuming we are required to calculate the new volume at the height of 4.05 km

We are going to use the combined gas law.

  • According to the combined gas law;

[tex]\frac{P1V1}{T1}=\frac{P2V2}{T2}[/tex]

  • Rearranging the formula;

[tex]V2=\frac{P1V1T2}{P2T1}[/tex]

[tex]V2=\frac{(759mmHg)(793L)(280.25K)}{(537mmHg)(290.05K)}[/tex]

[tex]V2=1082.96L[/tex]

Therefore, the new volume of the balloon at the height of 4.05 km is 1082.96 L

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