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Rama09 [41]
2 years ago
11

A sample of ideal gas at room temperature occupies a volume of 27.0 L at a pressure of 852 torr . If the pressure changes to 426

0 torr , with no change in the temperature or moles of gas, what is the new volume, V2
Chemistry
1 answer:
Lubov Fominskaja [6]2 years ago
5 0

The new volume V₂ is 5.4L

The new volume is calculated using Boyle's law formula

That  is  P₁V₁=P₂V₂

where,

p₁ =  852 torr

V₁ = 27.0

P₂ = 4260 torr

V₂=?

Make  V₂ the subject of the formula by diving into both sides of the equation by P₂

V₂ = P₁V₁/P₂

⇒ 852 *

V₂  is, therefore, = 5.4 L

Learn more about Boyle's law here: brainly.com/question/1696010

#SPJ4

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Nitric oxide is formed in automobile exhaust when nitrogen and oxygen in air react at high temperatures.N2(g) + O2(g) 2NO(g)The
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Answer : The correct option is, (E) 7.8 atm

Explanation :

The partial pressure of N_2 = 8.00 atm

The partial pressure of O_2 = 5.00 atm

K_p = 0.0025

The balanced equilibrium reaction is,

                               N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initial pressure     8.00      5.00            0

At eqm.               (8.00-x) (5.00-x)        2x

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{NO})^2}{(p_{N_2})(p_{O_2})}

Now put all the values in this expression, we get :

0.0025=\frac{(2x)^2}{(8.00-x)\times (5.00-x)}

By solving the terms, we get:

x=0.15atm

The equilibrium partial pressure of N_2 = (8.00 - x) = (8.00 - 0.15) = 7.8 atm

Therefore, the equilibrium partial pressure of N_2 is 7.8 atm.

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