Question

At \(T(K)\), the rms velocity of methane is \(x \, {ms}^{-1}\). What is the kinetic energy (in J) of 8 g of methane at the same temperature
(Assume methane as an ideal gas) 
 

• A. \(2 \times 10^{-3}x^2\)
• B. \(4 \times 10^{-3}x^2\)
• C. \(4 \times 10^{-3}x\)
• D. \(3 \times 10^{-3}x\)

Hint:

Remember the relationship between rms velocity, temperature, and molar mass. Also, the kinetic energy of an ideal gas is directly proportional to the number of moles and temperature.

Answer 1

The Correct Option is B

1. Molar mass of methane (CH\(_4\)): * Molar mass = 12 (C) + 4(1) (H) = 16 g/mol = 0.016 kg/mol 
2. Number of moles (n) of 8 g methane: n = mass / molar mass = 8 g / 16 g/mol = 0.5 moles 
3. Relationship between rms velocity (x) and temperature (T): \(x = \sqrt{\frac{3RT}{M}}\)
Squaring both sides: \(x^2 = \frac{3RT}{M}\)
Rearranging for RT: \(RT = \frac{Mx^2}{3}\)
4. Kinetic energy (KE) of n moles of an ideal gas: KE = \(\frac{3}{2} nRT\) 
5. Substitute the value of RT: KE = \(\frac{3}{2} n \left(\frac{Mx^2}{3}\right)\)
KE = \(\frac{1}{2} nMx^2\)
6. Substitute the values of n and M: KE = \(\frac{1}{2} \times 0.5 \, {mol} \times 0.016 \, {kg/mol} \times x^2\)
KE = \(0.004 \, {kg} \times x^2\)
KE = \(4 \times 10^{-3}x^2 \, {J}\)
Therefore, the kinetic energy of 8 g of methane at the same temperature is \(4 \times 10^{-3}x^2\) J. 
Final Answer: 
\(4 \times 10^{-3}x^2\) J.