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NEET PHYSICSEasy

The kinetic energy of one gram molecule of a gas at standard temperature and pressure is: (R = 8.31 J/mol-K)

A

0.56 × 10^4 J

B

1.3 × 10^2 J

C

2.7 × 10^2 J

D

3.4 × 10^3 J

Step-by-Step Solution

The kinetic energy (E) of 'one gram molecule' (which equals 1 mole) of an ideal gas is given by the formula E=32RTE = \frac{3}{2}RT. Given: R=8.31 J mol−1 K−1R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1} . Standard Temperature (STP) T=273 KT = 273 \text{ K} (\approx 0∘C0^\circ\text{C}). Substituting the values: E=32×8.31×273E = \frac{3}{2} \times 8.31 \times 273. E=1.5×2268.63=3402.9 JE = 1.5 \times 2268.63 = 3402.9 \text{ J}. In scientific notation, this is approximately 3.4×103 J3.4 \times 10^3 \text{ J}.

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