TS EAMCET · Chemistry · States of Matter

The kinetic energy in $J$ of $1$ mole of $N_{2}$ at $27 ° C$ is: $(R = 8.314 J {mol}^{- 1} K^{- 1})$

A $2, 494$
B $18706$
C $7482$
D $3741$

Verified Solution

Answer & Solution

Correct Answer

(D) $3741$

Step-by-step Solution

Detailed explanation

We know that kinetic energy is given by, KE=12mvrms2 And vrms=3RTm So, the average kinetic energy of 'n' moles of a gas is given by, KE=32nRT =32×1×8.314×(27+273) J =32×8.314×300 J =3741.3 J.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The kinetic energy in J of 1 mole of N2 at 27°C is: R=8.314 J mol-1 K-1

Step-by-step Solution

The kinetic energy in $J$ of $1$ mole of $N_{2}$ at $27 ° C$ is: $(R = 8.314 J {mol}^{- 1} K^{- 1})$