The kinetic energy of 1 g molecule of a gas at normal temperature and pressure is (R=8.31 J/mol-K)

Question

The kinetic energy of 1 g molecule of a gas at normal temperature and pressure is (R=8.31 J/mol-K)  (A)  3.4× 103J  (B)  2.97× 103J  (C)  1.2× 102J  (D)  0.66× 104J

Tardigrade Admin · Accepted Answer

E=(l/2)mv-2 =(1/2)M( (3RT/M) ) E=(3/2)RT E=(3/2)× 8.31× 273 E=3.4× 103J

Q. The kinetic energy of 1 g molecule of a gas at normal temperature and pressure is $(R = 8.31 J / m o l - K)$

$3.4 \times 10^{3} J$

$2.97 \times 10^{3} J$

$1.2 \times 10^{2} J$

$0.66 \times 10^{4} J$

$E = \frac{l}{2} m v^{- 2}$ $= \frac{1}{2} M (\frac{3 RT}{M})$ $E = \frac{3}{2} RT$ $E = \frac{3}{2} \times 8.31 \times 273$ $E = 3.4 \times 10^{3} J$

Q. The kinetic energy of 1 g molecule of a gas at normal temperature and pressure is (R=8.31J/mol−K)

3.4×103J

2.97×103J

1.2×102J

0.66×104J