The kinetic energy of one molecule of a gas at normal temperature and pressure will be (k=8.31 J mole K):

Question

The kinetic energy of one molecule of a gas at normal temperature and pressure will be (k=8.31 J mole K): (A)  1.7× 103 J (B)  10.2× 103 J (C)  3.4× 103 J (D)  6.8× 103 J

Tardigrade Admin · Accepted Answer

According to kinetic theory, K.E. of 1 g-mole of an ideal gas, E=(3/2) R T Hence, K.E. at normal temperature 0° C =273 K or E =(3/2) × 8.31 × 273=3.4 × 103 J

Q. The kinetic energy of one molecule of a gas at normal temperature and pressure will be ( $k = 8.31 J$ mole $K$ ):

$1.7 \times 1 0^{3} J$

$10.2 \times 1 0^{3} J$

$3.4 \times 1 0^{3} J$

$6.8 \times 1 0^{3} J$

According to kinetic theory, K.E. of $1 g$ -mole of an ideal gas, $E = \frac{3}{2} RT$
Hence, K.E. at normal temperature $0^{\circ} C$
$= 273 K$
or $E = \frac{3}{2} \times 8.31 \times 273 = 3.4 \times 1 0^{3} J$

Q. The kinetic energy of one molecule of a gas at normal temperature and pressure will be (k=8.31J moleK):

1.7×103J

10.2×103J

3.4×103J

6.8×103J