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  • Consider a container with a volume of 0.02 m3 that holds a gas at a temperature of 500 K. The gas has a molecular weight (Mm) of 28 g/mol. Calculate the root mean square velocity (v

Consider a container with a volume of 0.02 m3 that holds a gas at a temperature of 500 K. The gas has a molecular weight (Mm) of 28 g/mol. Calculate the root mean square velocity (vrms) of the gas molecules and the kinetic energy per mole of the gas.

Option: 1

6217.5 J/mol


Option: 2

5417.6 J/mol


Option: 3

4961.2 J/mol


Option: 4

7541.6 J/mol


Answers (1)

best_answer

Given data: Volume of container (V ) = 0.02 m3

Temperature (T) = 500 K

Molecular weight (Mm) = 28 g/mol

Boltzmann constant (k) = 1.38 × 10−23 J/K Avogadro’s number (NA) = 6.022 × 1023 mol−1

Step 1: Convert the molecular weight to kg/mol: Mm = 28 g/mol = 28 × 10−3 kg/mol

Step 2: Calculate the molar gas constant (R) using the Boltzmann constant (k) and Avogadro’s number (NA):

R = k × NA = 1.38 × 10−23 × 6.022 × 1023 = 8.31 J/(mol K)

Step 3: Calculate the root mean square velocity (vrms) using the formula:

v_{\mathrm{rms}}=\sqrt{\frac{3 \times R \times T}{M m}}

Substitute the given values and solve for vrms:

\begin{gathered} v_{\mathrm{rms}}=\sqrt{\frac{3 \times 8.31 \times 500}{28 \times 10^{-3}}} \\ v_{\mathrm{rms}} \approx 728.59 \mathrm{~m} / \mathrm{s} \end{gathered}

Step 4: Calculate the kinetic energy per mole of the gas (KE) using the formula:

K E=\frac{3}{2} \times R \times T

Substitute the given values and solve for KE:

\begin{gathered} K E=\frac{3}{2} \times 8.31 \times 500 \\ K E \approx 6217.5 \mathrm{~J} / \mathrm{mol} \end{gathered}

 

 

Posted by

Ajit Kumar Dubey

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