A $5 L$ $5 L$ container holds $16$ $16$ mol and $6$ $6$ mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from $320^{o} K$ $320^oK$ to $450^{o} K$ $450 ^oK$ . By how much does the pressure change?

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A $5 L$ $5 L$ container holds $16$ $16$ mol and $6$ $6$ mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from $320^{o} K$ $320^oK$ to $450^{o} K$ $450 ^oK$ . By how much does the pressure change?

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Answer 1 · 2016-04-21T16:35:37

$56.4 %$ $56.4%$ increase

Explanation:

Caution : Dont write $X^{o}$ $X^o$ if it is in kelvin, write $X K$ $X K$ straightaway.

Considering they are ideal gases, they follow

$P V = n R T$ $PV = nRT$

Here, $V$ $V$ is constant and $R$ $R$ is anyway constant, so we are left with

$\frac{P_{1}}{n_{1} T_{1}} = \frac{P_{2}}{n_{2} T_{2}}$ $P_1/(n_1 T_1)=P_2/(n_2 T_2)$

Note that, $\frac{16}{6} > \frac{4}{3}$ $16/6>4/3$ . So gas B gets totally consumed up. So, $\frac{6}{3} = 2$ $6/3 = 2$ moles of new has was formed, so in the process $8$ $8$ moles of gas A was taken up and $8$ $8$ moles of A is remaining. So in the reaction, total $n_{1} (= 16 + 6 = 22)$ $n_1 (=16+6 = 22)$ moles of gas is transformed into $n_{2} (= 2 + 8 = 10)$ $n_2(=2+8=10)$ moles of gas.

we are given,

$T_{1} = 320 K$ $T_1 = 320 K$
$T_{2} = 450 K$ $T_2 = 450K$

So, $\frac{P_{1}}{P_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}} = \frac{22 \cdot 320}{10 \cdot 450} \approx 1.564$ $P_1/P_2 = (n_1T_1)/(n_2T_2) = (22*320)/(10*450) ~~1.564$

So pressure is about $56.4 %$ $56.4%$ increased.

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A $5 L$ $5 L$ container holds $16$ $16$ mol and $6$ $6$ mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from $320^{o} K$ $320^oK$ to $450^{o} K$ $450 ^oK$ . By how much does the pressure change?

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Explanation:

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A 5 L5L5 L container holds 16 1616 mol and 6 66 mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from 320^oK320oK320^oK to 450 ^oK450oK450 ^oK. By how much does the pressure change?

1 Answer

Explanation:

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A $5 L$ $5 L$ container holds $16$ $16$ mol and $6$ $6$ mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from $320^{o} K$ $320^oK$ to $450^{o} K$ $450 ^oK$ . By how much does the pressure change?