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Boltzmann’s constant is 1.38066 × 10−23 J/K, and the universal gas constant is 8.31451 J/K · mol.
If 2.4 mol of a gas is confined to a 4.3 L vessel at a pressure of 11 atm, what is the average kinetic energy of a gas molecule? Answer in units of J.

Respuesta :

skyluke89

Answer:

[tex]4.95\cdot 10^{-21} J[/tex]

Explanation:

First of all, let's convert everything into SI units:

n = 2.4 mol (number of gas moles)

[tex]p=11 atm = 1.11\cdot 10^6 Pa[/tex] (gas pressure)

[tex]V=4.3 L=4.3\cdot 10^{-3} m^3[/tex] (gas volume)

R = 8.31451 J/K · mol (gas constant)

The ideal gas equation states that

[tex]pV=nRT[/tex]

Solving for T, we find the gas temperature

[tex]T=\frac{pV}{nR}=\frac{(1.11\cdot 10^6)(4.3\cdot 10^{-3})}{(2.4)(8.31451)}=239.2 K[/tex]

And now we can find the average kinetic energy of the gas:

[tex]E_K = \frac{3}{2}kT[/tex]

where

k = 1.38066 × 10−23 J/K is the Boltzmann's constant

Substituting,

[tex]E_K = \frac{3}{2}(1.38066\cdot 10^{-23} J/K)(239.2 K)=4.95\cdot 10^{-21} J[/tex]

The average kinetic energy of the gas molecule is 4.97 × 10⁻²¹ J

Calculating the average kinetic energy of a gas molecule

From the question, we are to calculate the average kinetic energy of the gas molecule.

The average kinetic energy of a gas molecule can be calculated from the formula,

[tex]KE_{avg}=\frac{3}{2}k_{b}T[/tex]

Where [tex]k_{b}[/tex] is the Boltzmann's constant

and T is the temperature

Now, we will determine the temperature of the gas

From the given information

R = 8.31451 J/K.mol

n = 2.4 moles

V = 4.3 L

P = 11 atm = 1114.58 KPa

From the ideal gas equation

PV=nRT

Then, we can write that

[tex]T = \frac{PV}{nR}[/tex]

Putting the parameters into the equation, we get

[tex]T = \frac{1114.58 \times 4.3}{2.4 \times 8.31451}[/tex]

T = 240.1772

T ≅ 240.2 K

Now, for the average kinetic energy of the gas molecule

[tex]KE_{avg}=\frac{3}{2}k_{b}T[/tex]

From the given information

[tex]k_{b} = 1.38066 \times 10^{-23}[/tex]

∴ [tex]KE_{avg}=\frac{3}{2} \times 1.38066 \times 10^{-23} \times 240.2[/tex]

[tex]KE_{avg}= 4.97 \times 10^{-21} \ J[/tex]

Hence, the average kinetic energy of the gas molecule is 4.97 × 10⁻²¹ J

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