Question

Average kinetic energy of one molecule of H2 at 77°C is (1) 1.25 × 10–22 J
(3) 7.25 × 10–21 J
(2) 3.25 × 10–19 J
(4) 4.1 × 10–22 J

Answer 1

Answer:

7.25*10^-21J

Explanation:

Answer 2

To find:

Average kinetic energy of one molecule of H2 at 77°C is ?

Calculation:

For "n" number of moles , the average kinetic energy is given as follows :

$\boxed{ \sf \therefore \: E = \dfrac{3}{2} nRT}$

For 1 molecule , the equation can be changed to :

$\sf \implies \: E = \dfrac{(\dfrac{3}{2} nRT)}{n\times avagadro \: no.}$

$\boxed{ \sf \implies \: E = \dfrac{(\dfrac{3}{2} RT)}{avagadro \: no.}}</p><p>$

Putting the values in SI UNITS :

$\sf \implies \: E = \dfrac{\dfrac{3}{2} \times 8.314 \times (77 + 273)}{6.023 \times {10}^{23} }</p><p>$

$\sf \implies \: E = \dfrac{\dfrac{3}{2} \times 8.314 \times 350}{6.023 \times {10}^{23} }</p><p>$

$\sf \implies \: E = \dfrac{3 \times 4.157 \times 350}{6.023 \times {10}^{23} }</p><p>$

$\sf \implies \: E = 7.2469 \times {10}^{ - 21} \: joule$

$\sf \implies \: E \approx 7.25 \times {10}^{ - 21} \: joule$

So, the average energy for that 1 molecule is

$\boxed{ \bf\: E \approx 7.25 \times {10}^{ - 21} \: joule}$