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A balloon is filled with a gas to a certain volume at a certain pressure at 0.987°C. If the pressure exerted on the balloon is doubled, what must the temperature (in °C) be to keep the balloon inflated at the same volume?

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Edufirst

Answer:

  • 275. °C

Explanation:

1) Data:

a) Constant volume

b) Initial pressure: P₀

c) Intitial temperature: T₀ = 0.987°C = 0.987 + 273.15 K = 274.137 K

d) Final pressure: P₁ = 2 P₀

e) Final temperature: unknown, T₁

2) Applied principles:

  • Gay - Lussac's law: at constant volume, the pressure and temperature of the gases are in direct proportion:

        ⇒ P / T = constante ⇒ P₁ / T₁ = P₀ / T₀

3) Solution:

a) Solve for T₁

  • T₁ = P₁ T₀ / P₀

b) Substitute the data:

  • T₁ = 2P₀ × 274.137 K / P₀ = 548.274 K

c) Converto to °C:

  • T₁ = 548.274 - 273.15 = 275.124 °C ≈ 275. °C (three significant figures)
mahima6824soni

The temperature (in °C)  to keep the balloon inflated at the given volume is  275. °C.

What is temperature?

Temperature is a measurement of degree that tells the coldness or hotness of any substance.

Given,

Temperature =0.987°C

converted into kelvin -0.987 + 273.15 K = 274.137 K

By Gay-Lussac's law

The pressure and temperature of the gases are in direct proportion at constant volume.

[tex]\bold{\dfrac{P}{T} = constants =\dfrac{P_1}{ T_1} = \dfrac{P_0}T_0} }[/tex]

[tex]\bold{T_1 ={P_1} \dfrac{T_0}P_0} }\\[/tex]

[tex]\bold{T_1 = 2P_0 \dfrac{274.137 K}{P_0} = 548.274 K }[/tex]

Converting kelvin to Celsius:

548.274 - 273.15 = 275.124 °C

Thus, the temperature is  275. °C.

Learn more about temperature, here:

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