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  • Consider a gas with a molecular weight (Mm) of 32 g/mol. The gas is at a temperature of 300 K. Calculate the root mean square velocity (vrsm) of the gas molecules.<div class='qna-opt

Consider a gas with a molecular weight (Mm) of 32 g/mol. The gas is at a temperature of 300 K. Calculate the root mean square velocity (vrsm) of the gas molecules.

Option: 1

541.65 m/s


Option: 2

450.32 m/s


Option: 3

469.59 m/s


Option: 4

516.88 m/s


Answers (1)

best_answer

Given data: Molecular weight (Mm) = 32 g/mol

Temperature (T) = 300 K

Boltzmann constant (k) = 1.38 × 10−23 J/K

Step 1: Convert the molecular weight to kg/mol: Mm = 32 g/mol = 32 × 10−3 kg/mol

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

R=k \times N_A=1.38 \times 10^{-23} \times 6.022 \times 10^{23}=8.31 \mathrm{~J} /(\mathrm{mol} \mathrm{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 300}{32 \times 10^{-3}}} \\ v_{\mathrm{rms}} \approx 516.88 \mathrm{~m} / \mathrm{s} \end{gathered}

Therefore, the option correct is (4).

Posted by

avinash.dongre

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