Question

A sample 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?

Answer 1

Answer: The final temperature would be 1250.7 K.

Explanation: We are given a sample of helium gas, the initial conditions are:

$V_{initial}=355mL=0.355L$  (Conversion factor: 1L = 1000 mL)

$T_{initial}=23\°C=296K$ (Conversion Factor: 1° C = 273 K)

The same gas is expanded at constant pressure, so the final conditions are:

$V_{initial}=1.50L$

$T_{initial}=?K$

To calculate the final temperature, we use Charles law, which states that the volume of the gas is directly proportional to the temperature at constant pressure.

$V\propto T$

$\frac{V_{initial}}{T_{initial}}=\frac{V_{final}}{T_{final}}$

Putting the values, in above equation, we get:

$\frac{0.355L}{296K}=\frac{1.50L}{T_{final}}$

$T_f=1250.7K$

Answer 2

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

Further explanation

There are several gas equations in various processes:

• 1. The general ideal gas equation

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

• 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

$\large{\boxed{\bold{\frac{V1}{V2}\:=\:\frac{n1}{n2} }}}$

• 3. Boyle's Law

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

$\large{\boxed{\bold{p1.V1=p2.V2}}}$

• 4. Charles's Law

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

$\large{\boxed{\bold{\frac{V1}{T1}=\frac{V2}{T2} }}}$

• 5. Gay Lussac's Law

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

$\large{\boxed{\bold{\frac{P1}{T1}=\frac{P2}{T2}}}}$

• 6. Law of Boyle-Gay-Lussac

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

$\large{\boxed{\bold{\frac{P1.V1}{T1}=\frac{P2.V2}{T2}}}}$

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

$\frac{V1}{T1}=\frac{V2}{T2}$

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

$\frac{355.10^{-3}}{296}=\frac{1.5}{T2}$

T2 = 1250, 704 K

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