Question

kinetic energy of 80g of methane gas at 227​

Answer 1

Answer:

please refers to the image given above

Answer 2

To find:

Kinetic energy of 80 grams of methane gas at 227 °C.

Calculation:

The average kinetic energy of any gas at a particular temperature is given as follows:

$\boxed{ \sf{KE = \dfrac{3}{2} \times K_{b} \times \theta}}$

Putting available values in SI unit:

$= > \sf{KE = \dfrac{3}{2} \times \dfrac{8.314}{6.023 \times {10}^{23} } \times \theta}$

$= > \sf{KE = \dfrac{3}{2} \times \dfrac{8.314}{6.023 \times {10}^{23} } \times (227 + 273)}$

$= > \sf{KE = \dfrac{3}{2} \times \dfrac{8.314}{6.023 \times {10}^{23} } \times 500}$

$= > \sf{KE = \dfrac{8.314}{6.023 \times {10}^{23} } \times 750}$

$= > \sf{KE = 1035.2 \times {10}^{ - 23} }$

$= > \sf{KE = 1.035\times {10}^{ - 20} \: joules}$

$= > \sf{KE = 1.035\times {10}^{ - 20} \times 4.5\: cal}$

$= > \sf{KE = 4.65\times {10}^{ - 20} \: cal}$

$= > \sf{KE = 4.65\times {10}^{ - 23} \: kcal}$

So, final answer is:

$\boxed{\bf{KE = 4.65\times {10}^{ - 23} \: kcal}}$