A $6 L$ $6 L$ container holds $4$ $4$ mol and $6$ $6$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $480^{o} K$ $480^oK$ to $270^{o} K$ $270^oK$ . How much does the pressure change?

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A $6 L$ $6 L$ container holds $4$ $4$ mol and $6$ $6$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $480^{o} K$ $480^oK$ to $270^{o} K$ $270^oK$ . How much does the pressure change?

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Answer 1 · 2017-11-12T23:27:55

Final Pressure : $P_{f} = (\frac{n_{f} T_{f}}{n_{i} T_{i}}) P_{i} = 0.3765$ $P_f = (\frac{n_fT_f}{n_iT_i})P_i = 0.3765$ $M P a$ $MPa$
Pressure Change :
$\DeltaP = P_f-P_i = [\frac{n_fT_f}{n_iT_i}-1]P_i = -2.9706$ $MPa$

Explanation:

Ideal Gas Equation of State: $PV=nRT$ ,

The volume $V$ is held constant;
$R=4.184 J/(mol.K)$ is the universal gas constant.

Everything else ( $P$ , $n$ and $T$ ) vary. So let us rewrite the EoS making this fact explicit by grouping all the terms that are held constant inside a parenthesis.

$P=(R/V)nT$ : terms inside the parenthesis are constant

( $P_i$ , $n_i$ , $T_i$ ) : Pressure, number of moles and temperature before the reaction.
( $P_f$ , $n_f$ , $T_f$ ) : Pressure, number of moles and temperature after the reaction.

$P_i = (R/V) n_iT_i; \qquad P_f = (R/V)n_fT_f; \qquad P_f/P_i = (n_fT_f)/(n_iT_i);$
$P_f = (\frac{n_fT_f}{n_iT_i})P_i; \qquad \DeltaP = P_f-P_i = [\frac{n_fT_f}{n_iT_i}-1]P_i$

Stoichiometry: $4A+6B \rightarrow 2A_2B_3$

$4$ mols of $A$ combine with $6$ mols of $B$ to give $2$ mols of $A_2B_3$

Before Reaction: $P_iV=(n_a+n_b)RT_i$
$n_a = 4$ $mols; \quad n_b = 6$ $mols; \qquad n_i=n_a+n_b=10$ $mols$
$T_i=480 K; \qquad V=6L=6\times10^{-3}m^3$
$P_i = (n_a+n_b)\frac{RT_i}{V} = 3.3472\times10^6$ $Pa = 3.3472$ $MPa$ .

After Reaction: $P_fV = n_{ab}RT_f$
$n_f = n_{ab} = 2$ $mols;$
$T_f=270 K; \qquad V=6L=6\times10^{-3}m^3$

$P_f = (\frac{n_fT_f}{n_iT_i})P_i = (\frac{2\times270K}{10\times480K})\times3.3472$ $MPa$
$\qquad= 0.3765$ $MPa$
$\DeltaP = P_f-P_i = [\frac{n_fT_f}{n_iT_i}-1]P_i = -2.9706$ $MPa$

The pressure change is negative, indicating that the pressure decreases.

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A $6 L$ $6 L$ container holds $4$ $4$ mol and $6$ $6$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $480^{o} K$ $480^oK$ to $270^{o} K$ $270^oK$ . How much does the pressure change?

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A 6 L6L6 L container holds 4 44 mol and 6 66 mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from 480^oK480oK480^oK to 270^oK270oK270^oK. How much does the pressure change?

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Explanation:

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A $6 L$ $6 L$ container holds $4$ $4$ mol and $6$ $6$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $480^{o} K$ $480^oK$ to $270^{o} K$ $270^oK$ . How much does the pressure change?