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  • An ideal gas is contained in a cubic container with sides of length 0.1 m. The gas is at a temperature of 400 K. Each gas molecule has a mass of <img alt="2.5 \times 10^{-26} \mathrm{~kg}" src

An ideal gas is contained in a cubic container with sides of length 0.1 m. The gas is at a temperature of 400 K. Each gas molecule has a mass of 2.5 \times 10^{-26} \mathrm{~kg}. Calculate the pressure exerted by the gas molecules on the walls of the container. Also, calculate the root mean square (RMS) speed of the gas molecules.

Option: 1

7.45 \times 10^{-19} \mathrm{~J}


Option: 2

8.28 \times 10^{-21} \mathrm{~J}


Option: 3

6.14 \times 10^{-17} \mathrm{~J}


Option: 4

4.25 \times 10^{-26} \mathrm{~J}


Answers (1)

best_answer

Given data: Side length of the container (a) = 0.1 m

Temperature (T) = 400 K

\text { Mass of each gas molecule }(m)=2.5 \times 10^{-26} \mathrm{~kg}

\text { Boltzmann constant }(k)=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}

Step 1: Calculate the RMS speed of the gas molecules using the formula:

v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}

Substitute the given values and solve for vrms:

\begin{gathered} v_{\mathrm{rms}}=\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 400}{2.5 \times 10^{-26}}} \\ v_{\mathrm{rms}} \approx 544.16 \mathrm{~m} / \mathrm{s} \end{gathered}

Step 2: Calculate the density of gas molecules (ρ) using the formula:

ρ = N/V

Substitute the given values and solve for ρ:

\begin{aligned} \rho & =\frac{N}{V}=\frac{1.5 \times 10^{24}}{(0.1)^3} \\ \rho & \approx 1.5 \times 10^{29} \mathrm{~m}^{-3} \end{aligned}

Step 3: Calculate the pressure (P) exerted by the gas molecules on the walls of the container using the formula:

P=\frac{2}{3} \rho v_{\mathrm{rms}}^2

Substitute the values of ρ and vrms and solve for P:

\begin{gathered} P=\frac{2}{3} \times\left(1.5 \times 10^{29}\right) \times(544.16)^2 \\ P \approx 8.79 \times 10^7 \mathrm{~Pa} \end{gathered}

Step 4: Calculate the average kinetic energy (Kavg) of a gas molecule using the formula:

K_{\mathrm{avg}}=\frac{3}{2} k T

Substitute the given values of temperature and k and solve for Kavg:

\begin{gathered} K_{\text {avg }}=\frac{3}{2} \times 1.38 \times 10^{-23} \times 400 \\ K_{\text {avg }} \approx 8.28 \times 10^{-21} \mathrm{~J} \end{gathered}

Therefore, the correct option is (2).

Posted by

shivangi.shekhar

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