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ch4aika [34]
3 years ago
14

A certain box contains 200 particles of an ideal gas. how many times more likely is it to find the particles evenly split betwee

n the left and right halves of the box, than to find 160 particles on one side and 40 on the other?
Chemistry
1 answer:
Strike441 [17]3 years ago
5 0

 Each particle has two possibilities: either it is on the left hand side or it is on the right hand side.  

The probability of having them equally split is  

P(100,100) = 200C100 / 2^200  

where 200C100 is the binomial coefficient.  

For the uneven distribution we have  

P= 2 * 200C160 / 2^200  

where the factor 2 comes in because we could have 160/40 or 40/160 as a division.  

So the ratio of probabilities is  

200C100 /(2 * 200C160 )  

= 2.2 * 10^16

Hope this helps!

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CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
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Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

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V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

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The balanced chemical reaction will be:

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Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

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Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

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