Asample of helium gas occupies 355ml at 23°c. if the container the he is in is expanded to 1.50 l at constant pressure, what is the final temperature for the he at this new volume?

The final temperature for the He at this new volume = 1250,704 liters

There are several gas equations in various processes:

PV = nRT

PV = NkT

N = number of gas particles

n = number of moles

R = gas constant (8,31.10³ J / kmole K)

k = Boltzmann constant (1,38.10⁻²³)

n = = N / No

n = m / M

n = mole

No = Avogadro number (6.02.10²³)

m = mass

M = relative molecular mass

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

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

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

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

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

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

In the problem, the conditions that are set constant are Pressure, so we use Charles' Law

We first convert the known number

V1 = 355 ml = 355.10-3 liters

T1 = 23 C = 23 + 273 = 296 K

V2 = 1.5 liters

We enter the formula

T2 = 1250, 704 K

a description of Charles’s law

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Charles's law

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

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your answer would be “dry”