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xenn [34]
3 years ago
12

What is the volume of the air in a balloon that occupies 8.7 L at 50.0°C if the temperature is lowered to 25.0°C?

Chemistry
1 answer:
dangina [55]3 years ago
3 0

Answer:

8.03L

Explanation:

Given parameters:

Initial volume, V₁  = 8.7L

Initial temperature T₁  = 50°C , in Kelvin gives 50+273  = 323K

Final temperature T₂ = 25°C, in Kelvin gives 25 + 273  = 298K

Unknown:

Final volume V₂  = ?

Solution:

Since we are dealing with volume and temperature relationships under constant pressure, we simply apply the Charles's law.

Charles's law states that "the volume of a given mass of gas is directly proportional to the temperature provided the pressure remains constant".

Mathematically, we have;

              \frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }

where V andV T are volume and temperature, 1 and 2 are initial and final states.

Input the parameters and solve for V₂;

            \frac{8.7}{323}  =  \frac{V_{2} }{298}

          V₂  = 8.03L

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<u>Answer:</u>

<em>4.5 L water we have in litres (L).</em>

<em><u></u></em>

<u>Explanation:</u>

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\Delta T = Final T - Initial T

Q is the heat energy in calories

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Plugging in the values  

\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$

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True

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