Assertion (A) At 300 K,kinetic energy of 16 | vedclass

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(A) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = \frac{3}{2} nRT$.For $16 \ g$ of methane ($CH_4$,molar mass = $16 \ g/mol

Vedclass · Accepted Answer

Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$

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(A) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = \frac{3}{2} nRT$.For $16 \ g$ of methane ($CH_4$,molar mass = $16 \ g/mol$),moles $n_1 = \frac{16}{16} = 1 \ mol$.For $32 \ g$ of oxygen ($O_2$,molar mass = $32 \ g/mol$),moles $n_2 = \frac{32}{32} = 1 \ mol$.Since both have $1 \ mol$ at the same temperature $(300 \ K)$,their kinetic energies are equal. Thus,$(A)$ is true.Reason $(R)$ states that at constant temperature,kinetic energy of one mole of all gases is equal,which is true $(KE = \frac{3}{2} RT)$.Therefore,$(R)$ is the correct explanation of $(A)$.

Assertion $(A)$ At $300 \ K$,kinetic energy of $16 \ g$ of methane is equal to the kinetic energy of $32 \ g$ of oxygen.
Reason $(R)$ At constant temperature,kinetic energy of one mole of all gases is equal.

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Assertion $(A)$ At $300 \ K$,kinetic energy of $16 \ g$ of methane is equal to the kinetic energy of $32 \ g$ of oxygen.Reason $(R)$ At constant temperature,kinetic energy of one mole of all gases is equal.