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Chapter 13: Problem 61

What is the total translational kinetic energy of the gas molecules of 0.420 mol of air at atmospheric pressure that occupies a volume of $1.00 \mathrm{L}\left(0.00100 \mathrm{m}^{3}\right) ?$

Short Answer

Expert verified
Answer: The total translational kinetic energy of the gas molecules of 0.420 mol of air at atmospheric pressure that occupies a volume of 1.00 L is approximately 5443.75 J.

Step by step solution

01

Determine temperature using the Ideal Gas Law

The Ideal Gas Law is given by the formula \(PV = nRT\). Since it's at atmospheric pressure, we can assume the pressure \(P = 1 \mathrm{atm} = 101325 \mathrm{Pa}\). We are given the number of moles of air, \(n = 0.420 \mathrm{mol}\), and the volume, \(V = 0.00100 \mathrm{m}^3\). The gas constant, \(R = 8.314 J/(mol \cdot K)\). Rearrange the Ideal Gas Law to solve for temperature (\(T\)): \(T = \dfrac{PV}{nR} = \dfrac{101325 \;\mathrm{Pa} \cdot 0.00100 \;\mathrm{m}^3}{0.420 \;\mathrm{mol} \cdot 8.314 \;J/(mol \cdot K)}\)
02

Calculate the temperature

Now, substitute the values we have into the equation to find the temperature: \(T = \dfrac{101325 \;\mathrm{Pa} \cdot 0.00100 \;\mathrm{m}^3}{0.420 \;\mathrm{mol} \cdot 8.314 \;J/(mol \cdot K)} = 292.11 \;K\)
03

Calculate the total translational kinetic energy

According to the Kinetic Theory of Gases, the total translational kinetic energy of a gas is given by the formula: \(E_{\text{trans}} = \dfrac{3}{2} nRT\) Now, substitute the values we found for \(n\), \(R\), and \(T\) into this formula: \(E_{\text{trans}} = \dfrac{3}{2} \cdot 0.420 \;\mathrm{mol} \cdot 8.314 \;J/(mol \cdot K) \cdot 292.11 \;K\)
04

Calculate the total translational kinetic energy

Finally, perform the calculation to find the total translational kinetic energy: \(E_{\text{trans}} = \dfrac{3}{2} \cdot 0.420 \;\mathrm{mol} \cdot 8.314 \;J/(mol \cdot K) \cdot 292.11 \;K = 5443.75 \;J\) The total translational kinetic energy of the gas molecules of 0.420 mol of air at atmospheric pressure that occupies a volume of \(1.00 L\) is approximately \(5443.75 J\).

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