contestada

A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5.00 L, and the pressure is 760. mmHg. The balloon ascends to an altitude of 20 km, where the pressure is 76.0 mmHg and the temperature is −50. ∘C. What is the new volume, V2, of the balloon in liters, assuming it doesn't break or leak?

Respuesta :

baraltoa

Answer:

38.05 L

Explanation:

To solve this problem use the formula:

                   V1P1T2 = V2P2T1===== V2 = V1P1T2/P1T1

where

V1: 5.00 L

P1: 760 mmHg

T1: 20 +273 : 293K  (remember to convert ºC to K)

V2: ?

P2: 76 mmHg

T2: -50 + 273 : 223 K

V2= 5.00L * 760 mmHg * 223 K /( 76 mmHg *293 )

V2 38.05 L

ardni313

The new volume, V2, of the balloon : 38.0546

Further explanation

There are several gas equations in various processes:

  • 1. The ideal ideal gas equation

[tex]\rm PV=nRT[/tex]

PV = NkT

N = number of gas particles

n = number of moles

R = gas constant (8,31.10 ^ 3 J / kmole K

k = Boltzmann constant (1,38.10⁻²³)

n = N / No

n = mole

No = Avogadro number (6.02.10²³)

n = m / m

m = mass

M = relative molecular mass

  • 2. Avogadro's hypothesis

In the same temperature and pressure, in the same volume conditions, the gas contains the same number of molecules

So it applies: the ratio of gas volume will be equal to the ratio of gas moles

[tex]\rm V1:V2=n1:n2[/tex]

  • 3. Boyle's Law

At a constant temperature, the gas volume is inversely proportional to the pressure applied

[tex]\rm p1.V1=p2.V2[/tex]

  • 4. Charles's Law

When the gas pressure is kept constant, the gas volume is proportional to the temperature

[tex]\rm \dfrac{V1}{T1}=\dfrac{V2}{T2}[/tex]

  • 5. Gay Lussac's Law

When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature

[tex]\rm \dfrac{P1}{T1}=\dfrac{P2}{T2}[/tex]

  • 6. Law of Boyle-Gay-Lussac

Combined with Boyle's law and Gay Lussac's law

[tex]\rm \dfrac{P1.V1}{T1}=\dfrac{P2.V2}{T2}[/tex]

P1 = initial gas pressure (N / m² or Pa)

V1 = initial gas volume (m³)

P2 = gas end pressure

V2 = the final volume of gas

T1 = initial gas temperature (K)

T2 = gas end temperature

The initial condition of the balloon:

volume of 5.00 L

the pressure 760. mmHg

temperature 20. ∘C = 20 + 273 = 293 K

baloon conditions after 20 km

the pressure 76.0 mmHg

temperature −50. ∘C. = -50 +273 = 223K

we use Boyle-Gay-Lussac Law

[tex]\rm \dfrac{P1V1}{T1}=\dfrac{P2.V2}{T2}\\\\\dfrac{760.5}{293}=\dfrac{76.V2}{223}\\\\V2=\boxed{\bold{38.0546\:L}}[/tex]

Learn more

a description of Charles’s law

brainly.com/question/5056208

Charles's law

brainly.com/question/9510865

State Boyle's, Charles's, and Gay-Lussac's laws

brainly.com/question/980439

Otras preguntas