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SashulF [63]
2 years ago
5

Two containers, X and Y, are each filled by an ideal gas at the same temperature. The volume of Y is half the volume of X. The n

umber of moles of gas in Y is three times the number of moles of the gas in X. The pressure of the gas in X is PX and the pressure of the gas in Y is PY. What is the ratio LaTeX: \frac{P_X}{P_Y}P Y P Y
Mathematics
1 answer:
-Dominant- [34]2 years ago
6 0

Answer:

The answer to the question is

The ratio of the two gas pressures   \frac{P_{x} }{P_{y} } , that is Px to Py = 1/6

Step-by-step explanation:

Let the gases Volumes be V₁ and V₂

Where volume of X = V₁ and

volume of Y = V₂

The volume of Y is half the volume of X

∴ V₂  =  \frac{1}{2} × V₁

Let the number of moles be n₁ and n₂ in X and Y respectively

therefore  n₂ = 3 × n₁

The pressure of the gas in X is Pₓ and the pressure of the gas in Y is  P_{y} then we have

P₁ × V₁  = n₁ × R × T₁ , and P₂ × V₂ = n₂ × R × T₂

(P₁ × V₁)/(n₁ × T₁) = (P₂ × V₂)/(n₂ × T₂)

but T₁ = T₂

Therefore

(P₁ × V₁)/n₁ = (P₂ × V₂)/n₂.  However  n₂ = 3 × n₁  and V₂  =  \frac{1}{2} × V₁ therefore substituting in the equation we have

(P₁ × V₁)/n₁ = (P₂ ×  \frac{1}{2} × V₁ )/(3 × n₁) from where

P₁ /P₂ =  (\frac{1}{2} × V₁ × n₁)/(V₁×3 × n₁) =0.5/3 = 1/6

The ratio of \frac{P_{x} }{P_{y} } = 1/6

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Joshua started cycling at 5:15 pm By 8:09 pm, he has covered a distance of 7250 mWhat was Joshua's average speed during that tim
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The average speed of Joshua during that time is 2500 m/h.

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It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.

The total time taken by Joshua from 5:15 pm to 8:09 pm is

2 { h } 54 \text { min }=2 \mathrm{h}+\frac{54}{60} {h}

Dividing we get,

2 { h } 54 \text { min }=2 \mathrm{h}+0.9 {h}

Adding, we have,

2 { h } 54 \text { min }=2.9 {h}

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