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  • Consider a container filled with an ideal gas at a temperature of 300 K. The container has a volume of 0.1 . The gas con

Consider a container filled with an ideal gas at a temperature of 300 K. The container has a volume of 0.1 m^3. The gas contains 1.0 × 10^{24} gas molecules.
Each molecule has a mass of 2.0 × 10^{-26} kg.
Calculate the pressure exerted by the gas on the walls of the container, based on the given assumptions.

Option: 1

1.46 × 10−20 Pa


 


Option: 2

1.36 × 10−22 Pa


Option: 3

2.45 × 10−12 Pa


Option: 4

1.47 × 10−18 Pa


Answers (1)

best_answer

Given data:

Temperature (T) = 300 K

Volume (V ) = 0.1 m^3
Number of molecules (N) = 1.0 × 1010^{24}
Mass of each molecule (m) = 2.0 × 10^{-26} kg
Boltzmann constant (k) = 1.38 × 10^{-23} J/K

We can use the ideal gas law to relate the pressure (P), volume (V ), number of molecules (N), and temperature (T):
                                               P V = N kT

First, let’s find the total kinetic energy (KE) of the gas molecules using the given temperature:

                                            K E=\frac{3}{2} N k T

Where the factor  accounts for the three degrees of freedom per molecule in 3D motion.
Substitute the value of KE into the equation for P V :

                                             P V=\frac{2}{3} K E

Now, we’ll calculate KE and then substitute it into the equation to find the pressure P:

\begin{aligned} K E & =\frac{3}{2} N k T \\ & =\frac{3}{2} \cdot 1.0 \times 10^{24} \cdot 1.38 \times 10^{-23} \cdot 300 \\ & =6.21 \times 10^{-2} \mathrm{~J} \end{aligned}

Substitute KE into P V = KE:

P V=\frac{2}{3} \cdot 6.21 \times 10^{-2} \mathrm{~J}=4.14 \times 10^{-2} \mathrm{~J}

Pressure is defined as force per unit area. Since we’re assuming no inter-molecular forces and no gravitational effects, the force is simply the change in momentum due to collisions. In a single collision, the change in momentum(?p) is 2m · v, where m is the mass of a molecule and v is its velocity.

Since the average velocity in each direction is zero, the average magnitude of velocity ( ?v) is also zero. However, the average squared magnitude of velocity

\left(\bar{v}^2\right) can be calculated from the temperature using the formula \bar{v}^2 =\frac{3 k T}{m}

\begin{aligned} \bar{v}^2 & =\frac{3 k T}{m} \\ & =\frac{3 \cdot 1.38 \times 10^{-23} \cdot 300}{2 \times 10^{-26}} \\ & =8.31 \times 10^2 \mathrm{~m}^2 / \mathrm{s}^2 \end{aligned}

Now, we can calculate the change in momentum (?p) per collision:

\begin{aligned} \Delta p & =2 m \cdot \bar{v} \\ & =2 \cdot 2 \times 10^{-26} \cdot \sqrt{8.31 \times 10^2} \\ & =2.92 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \end{aligned}

Finally, we can calculate the pressure P as the rate of change of momentum per unit area:

                                         P=\frac{\Delta p}{A}

The area A is the area of the container’s walls. Assuming a cube-shaped container, all sides have the same area, so we’ll consider one of the sides. Let’s say the length of each side is L, then A = L^2.
For a cubic container with a volume of V = L^3, given that V = 0.1 m^3, we can solve for L:

L=\sqrt[3]{V}=\sqrt[3]{0.1} \mathrm{~m} \approx 0.464 \mathrm{~m}

Now we can calculate the area A:

A=L^2=(0.464)^2 \mathrm{~m}^2 \approx 0.215 \mathrm{~m}^2

Substitute ?p and A into the pressure formula:

P=\frac{2.92 \times 10^{-24}}{0.215} \mathrm{~N} / \mathrm{m}^2 \approx 1.36 \times 10^{-22} \mathrm{~N} / \mathrm{m}^2

Since 1 N/m^2 = 1 Pa, the pressure is approximately 1.36 × 10^{-22} Pa.
Therefore, the correct option is 2.

Posted by

Devendra Khairwa

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