Question

40. If the velocity of gas (A), having molecular weight
(90), corresponding to the maxima of given velocity
graph at a given temperature is 300 m/s. Then
find value (to nearest integer) of kinetic energy
(in kilojoule) of 360 gram of gas "A".
Fraction
of molecule
Velocity
inor at 27°C​

Answer 1

Answer:

Your answer as follows:

The velocity corresponding to a gas "X" (MW=100) corresponding to the maxima of the curve given below at a given temperature is 200 m/s.Then kinetic energy of 300 gram of h

Explanation:

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Answer 2

Given:

• The velocity of gas (A) has a molecular weight 90g.
• The Maxima of a given velocity graph at a given temperature is 300 m/s.
• weight of gas (A) is 360g.

To find: kinetic energy of gas A?

Explanation:

as we have,

                         molecular weight (MW) = 90g

                                           velocity (Vp) = 300m/s

                                  weight of gas (W) = 360g

we konw that,

                                       $V_{p} = \sqrt{\frac{2RT}{M.W.} } = 300$

         ⇒                         $\frac{RT}{M.W.} = \frac{(300)^2}{2}$

          Kinetic energy  = $\frac{3}{2} nRT = \frac{3}{2}$×$\frac{W}{M.W.}$×$RT$

          kinetic energy  = $\frac{3}{2}$×$\frac{360}{2}$×$300$×$300$

                             K.E. = 24300000J

                                    = 24300KJ

Hence kinetic energy of 360g of gas "A" is 24300KJ