An $8 L$ $8 L$ container holds $8$ $8$ mol and $9$ $9$ mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from $370^{o} K$ $370^oK$ to $425^{o} K$ $425 ^oK$ . How much does the pressure change by?

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An $8 L$ $8 L$ container holds $8$ $8$ mol and $9$ $9$ mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from $370^{o} K$ $370^oK$ to $425^{o} K$ $425 ^oK$ . How much does the pressure change by?

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Answer 1 · 2017-04-13T12:09:20

30.377atm at 370K
34.8925atm at 425K

Explanation:

So first we need to know the moles of gas $A B_{2}$ $AB_2$ formed

So this is chemistry stoichiometry

$2 B + A = A B_{2}$ $2B + A = AB_2$

So the reagent in excess is A and some amount of it will remain till the end

So if $\frac{1mol of A B_{2}}{2mol of B}$ $("1mol of "AB_2)/"2mol of B"$ thus

$\frac{x mol of A B_{2}}{9mol of B}$ $("x mol of "AB_2)/"9mol of B"$

x = 4.5mol of $A B_{2}$ $AB_2$

Since $\frac{1 mol of A B_{2}}{1 mol of A}$ $("1 mol of "AB_2)/"1 mol of A"$

So the amount of AB2 is equal to the amount of A used

So A used is 4.5mol

But left out is 8 - 4.5 = 3.5mol

So mol of A + mol of AB2 =

4.5mol + 3.5mol = 8mol

Now calculate the volume of gas at 370K and 425K

using the formula

$P V = n R T$ $color(red)(PV = nRT)$

Where P is pressure in atm
n is moles
R is the universal gas constant which is equal to 0.0821L
T is temperature in kelvin
V is volume

Plug in the variables

$8 \times P = 8 m o l \cdot 0.0821 L \cdot 370^{\circ} K$ $8xxP = 8mol * 0.0821L * 370^@K$

$P = 0.0821 L \times 370^{\circ} K = 30.377 a t m$ $P = 0.0821L xx370^@K = 30.377atm$

Now calculate the pressure at different temperature

$8 \times P = 8 m o l \cdot 0.0821 L \cdot 425^{\circ} K$ $8xxP = 8mol * 0.0821L * 425^@K$

$P = 0.0821 L \times 425^{\circ} K = 34.8925 a t m$ $P = 0.0821L xx425^@K = 34.8925atm$

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An $8 L$ $8 L$ container holds $8$ $8$ mol and $9$ $9$ mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from $370^{o} K$ $370^oK$ to $425^{o} K$ $425 ^oK$ . How much does the pressure change by?

1 Answer

Explanation:

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An 8L8 L container holds 88 mol and 99 mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from 370oK370^oK to 425oK425 ^oK. How much does the pressure change by?

1 Answer

Explanation:

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An $8 L$ $8 L$ container holds $8$ $8$ mol and $9$ $9$ mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from $370^{o} K$ $370^oK$ to $425^{o} K$ $425 ^oK$ . How much does the pressure change by?