A weather balloon is inflated to a volume of 28.1 L at a pressure of 745 mmHg and a temperature of 29.0 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 380. mmHg and the temperature is -16.3 ∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

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Answer:

[tex]V_2=46.8 L[/tex]

Explanation:

For these changes in pressure and temperature, assuming that the gases in the balloon are ideal, we can apply the general equation of gases:

[tex]\frac{P_1*V_1}{T_1}=\frac{P_2*V_2}{T_2}[/tex]

Been 1 the initial conditions and 2 the final ones:

[tex]\frac{745 mmHg*28.1L}{302K}=\frac{380 mmHg*V_2}{256.7K}[/tex]

[tex]V_2=\frac{745 mmHg*28.1L*256.7K}{380 mmHg*302K}[/tex]

[tex]V_2=46.8 L[/tex]

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