An $8 L$ $8 L$ container holds $6$ $6$ mol and $15$ $15$ mol of gasses A and B, respectively. Every three of molecules of gas B bind to one molecule of gas A and the reaction changes the temperature from $310^{o} K$ $310^oK$ to $150^{o} K$ $150 ^oK$ . By how much does the pressure change?

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Answer 1 · 2018-03-06T13:00:57

There was a change in $57.573 atm$ $57.573"atm"$

We need to find $DeltaP$ , and by extension, $P_1$ and $P_2$ .

According to the Ideal Gas Law:

$PV=nRT$ . Rearranging to solve for $P$ :

$P=(nRT)/V$

Here, $V=8"L"$ , $n_1=21"mol"$ , $T_1=310"K"$ and $R=0.0821"L atm K"^-1"mol"^-1$ . Inputting:

$P_1=(21*0.0821*310)/8$

$P_1=66.809"atm"$

We now write down the reaction which occurred:

$6"A"+15"B"rarr5"AB"_3+"A"$

So we know that $n_2=6"mol"$ . Here, $T_2=150"K"$ . Inputting these two new values into the Ideal Gas Law:

$P_2=(6*0.0821*150)/8$

$P_2=9.236"atm"$

We know that $DeltaP=P_2-P_1$ . So here,

$DeltaP=9.236-66.809=-57.573"atm"$

An 8 L8L8 L container holds 6 66 mol and 15 1515 mol of gasses A and B, respectively. Every three of molecules of gas B bind to one molecule of gas A and the reaction changes the temperature from 310^oK310oK310^oK to 150 ^oK150oK150 ^oK. By how much does the pressure change?