The kinetic energy of one gram mole of a gas at normal temperature and pressure is (R = 8.31 J/mole-K)

The correct option is 3.4x10³ J.

To find the kinetic energy of one gram mole of an ideal gas at normal temperature and pressure (NTP), we can use the formula for the translational kinetic energy of one mole of a gas:

$E=\frac{3}{2}RT$
where:
• $E$ is the kinetic energy of the gas,
• $R$ is the universal gas constant ($R=8.31\text{J/mol·K}$), and
• $T$ is the absolute temperature in Kelvin.

At Normal Temperature and Pressure (NTP), the standard temperature is:

$T=273\text{K}$

Substituting the given values into the kinetic energy equation:

$E=\frac{3}{2}\times 8.31\times 273$

$E=1.5\times 8.31\times 273$

$E=12.465\times 273$

$E\approx 3402.9\text{J}$

Expressing the result in scientific notation gives:

$E\approx 3.4\times {10}^3\text{J}$