Question

The total kinetic energy in Joules of the molecules in $8$ grams of $C{H}_4$ at ${27}^{\circ }C$.

• A. $18.7065$
• B. $187.065$
• C. $1870.65$
• D. $1.87065$

Answer 1

The correct option is C $1870.65$

Total kinetic energy $=n\left(\frac{3}{2}RT\right)$

where n = number of moles of the gas

R = Gas constant

T = Absolute temperature

Molecular weight of methane,

$C{H}_4=12+4\times 1=16$

Number of moles of methane in 8.0 gm of methane

$=\frac{8.0}{16.0}=0.5$

R = 8.314 joules/K/mole

$T=27+273=300K$

The total kinetic energy of the molecules in 8.0 gm of methane at ${27}^{o}C=n\times \frac{3}{2}RT$

$=0.5\times \frac{3}{2}\times 8.314\times 300=1870.65$ joules

Hence, the correct option is $\text{C}$