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ludmilkaskok [199]

2 years ago

8

A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r

ises in the atmosphere to an altitude where the pressure is 385 mmHgmmHg and the temperature is -14.8 ∘C∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

1 answer:

elena-s [515]2 years ago

5 0

Answer:

43.96 L

Explanation:

We are given that

$V_1=26.8 L$

$P_1=744mm Hg$

$T_1=31.2^{\circ} C=31.2+273=304.2K$

$P_2=385mmHg$

$T_2=-14.8^{\circ}=273-14.8=258.2K$

We know that

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1T_2}{P_2T_1}$

Substitute the values

$V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}$

$V_2=43.96 L$

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

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Question 17 A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off

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Complete Question:

A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm.

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(Question attached)

Answer:

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Explanation:

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When the system reaches equilibrium the iron and water will be the same temperature, $T_{e}$ . The energy lost from the iron will be equal to the energy gained by the water. It is assumed that the only heat exchange is between the iron and water and no exchange with the surroundings.

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$-Q_{iron}=Q_{water}$ (Eq 2)

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$m_{water}=100.0 g, c_{water}=4.186 J/kg.\°C, T_{initial,water}=23 \°C, T_{e}=27.6 \°C$

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$m_{iron}=59.1 g, c_{iron} = ? J/kg.\°C, T_{initial,iron}=85 \°C, T_{e}=27.6 \°C$

Substituting Eq 1 into Eq 2 and details extracted from the question:

$-m_{iron}c_{iron}(T_{iron,e}-T_{initial})=m_{water}c_{water}(T_{water,e}-T_{initial})$

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Explanation:

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