A $4 L$ $4 L$ container holds $4$ $4$ mol and $4$ $4$ mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from $280^{o} K$ $280^oK$ to $320^{o} K$ $320^oK$ . By how much does the pressure change?

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A $4 L$ $4 L$ container holds $4$ $4$ mol and $4$ $4$ mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from $280^{o} K$ $280^oK$ to $320^{o} K$ $320^oK$ . By how much does the pressure change?

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Answer 1 · 2017-04-19T04:46:32

$10 a t m$ $10 atm$

Explanation:

First, you need to turn the information into a chemical equation:
$3 A (g) + 4 B (g) \to A_{3} B_{4} (g)$ $3A(g) + 4B(g) -> A_3B_4(g)$

Then you find the moles of product formed and moles of reactant left. You should get 1 mol of $A_{3} B_{4} (g)$ $A_3B_4(g)$ and 1 mol $A (g)$ $A(g)$ left.

After that you just input the information into the formula $P_{T o t a l} = \frac{n_{T o t a l} R T}{V}$ $P_(Total) = (n_(Total)RT)/V$

$P_{T o t a l} = \frac{2 m o l s \cdot 0.08206 L atm m o l^{- 1} K^{- 1} \cdot 320 K}{4 L}$ $P_(Total) = (2mols * 0.08206 "L atm " mol^-1 K^-1 * 320 K)/(4L)$
$P_{T o t a l} = 13.1296 a t m$ $P_(Total) = 13.1296atm$

Since you only have one significant figure, the answer would be $10 a t m$ $10 atm$ .

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A $4 L$ $4 L$ container holds $4$ $4$ mol and $4$ $4$ mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from $280^{o} K$ $280^oK$ to $320^{o} K$ $320^oK$ . By how much does the pressure change?

1 Answer

Explanation:

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A 4 L4L4 L container holds 4 44 mol and 4 44 mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from 280^oK280oK280^oK to 320^oK320oK320^oK. By how much does the pressure change?

1 Answer

Explanation:

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A $4 L$ $4 L$ container holds $4$ $4$ mol and $4$ $4$ mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from $280^{o} K$ $280^oK$ to $320^{o} K$ $320^oK$ . By how much does the pressure change?