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hjlf
3 years ago
11

A weather balloon is inflated to a volume of 27.3 L at a pressure of 738 mmHg and a temperature of 26.9 ∘C. The balloon rises in

the atmosphere to an altitude where the pressure is 375 mmHg and the temperature is -15.6 ∘C.
Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.
Chemistry
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

The new volume of the balloon is 46.1 L

Explanation:

Step 1: Data given

Initial volume of the balloon = 27.3 L

Initial pressure in the balloon = 738 mmHg = 0.97105 atm

Initial temperature in the balloon = 26.9 °C = 300.05 K

The pressure decreases to 375 mmHg = 0.493421 atm

The temperature lowers to -15.6 °C = 257.55 K

Step 2: Calculate the volume

P1*V1 / T1 = P2 *V2 / T2

⇒with P1 = the Initial pressure = 738 mmHg = 0.97105 atm

⇒with V1 = Initial volume of the balloon = 27.3 L

⇒with T1 = Initial temperature in the balloon = 26.9 °C = 300.05 K

⇒with P2 = the decreased pressure = 0.493421 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the lowered temperature = 257.55 K

0.97105 * 27.3 / 300.05 = 0.493421*V2 / 257.55

V2 = 46.1 L

The new volume of the balloon is 46.1 L

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The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
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Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
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Q = mL

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Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

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Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

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It is given in the question that

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