Question

If the kinetic energy of the molecules in \(8.0\,\text{g}\) methane at \(47^\circ\text{C}\) is \(x\,\text{kJ}\), then the kinetic energy (in kJ) of the molecules in the same amount of dihydrogen at \(127^\circ\text{C}\) is

• A. \(x\)
• B. \(5x\)
• C. \(10x\)
• D. \(20x\)

Hint:

For an ideal gas, \[ \boxed{ K=\frac32nRT. } \] Therefore, \[ \boxed{ K\propto nT. } \] Always convert temperature into Kelvin before substitution.

Answer 1

The Correct Option is C

Step 1: Use the expression for total kinetic energy of a gas. The total kinetic energy is \[ K=\frac32nRT, \] where \(n\) is the number of moles. Thus, \[ K\propto nT. \]

Step 2: Calculate the number of moles. For methane, \[ M_{\mathrm{CH_4}}=16\,\text{g mol}^{-1}, \] \[ n_1=\frac{8}{16}=\frac12. \] Temperature, \[ T_1=47+273=320\,\text{K}. \] For hydrogen, \[ M_{\mathrm{H_2}}=2\,\text{g mol}^{-1}, \] \[ n_2=\frac{8}{2}=4. \] Temperature, \[ T_2=127+273=400\,\text{K}. \]

Step 3: Find the ratio of kinetic energies. \[ \frac{K_2}{K_1} = \frac{n_2T_2}{n_1T_1} = \frac{4\times400}{\frac12\times320} = 10. \] Hence, \[ K_2=10K_1=10x. \] Therefore, \[ \boxed{10x.} \] Hence, the correct option is \(\boxed{(C)}\).