A $5 L$ $5 L$ container holds $8$ $8$ mol and $10$ $10$ mol of gasses A and B, respectively. Every five of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from $360^{o} K$ $360^oK$ to $270^{o} K$ $270 ^oK$ . By how much does the pressure change?

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A $5 L$ $5 L$ container holds $8$ $8$ mol and $10$ $10$ mol of gasses A and B, respectively. Every five of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from $360^{o} K$ $360^oK$ to $270^{o} K$ $270 ^oK$ . By how much does the pressure change?

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Answer 1 · 2018-02-02T21:10:53

The total pressure $P$ $P$ inside the $V = 5 L$ $V = 5 L$ container is due to the pressures of both gasses $A$ $A$ and $B$ $B$ :

$P = P_{A} + P_{B}$ $P = P_A + P_B$ .

According to the ideal gas law:

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

If we consider gasses $A$ $A$ and $B$ $B$ to be ideal, then this equation must apply to each one of them. For the total pressure, this means that:

$P = \frac{n_{A} R T_{A}}{V_{A}} + \frac{n_{B} R T_{B}}{V_{B}}$ $P = (n_ART_A)/V_A + (n_BRT_B)/V_B$ .

If we also assume that both gasses can occupy all the container's volume and that they will always be at thermal equilibrium with each other, then:

$P = \frac{R T}{V} (n_{A} + n_{B})$ $P = (RT)/V(n_A + n_B)$ .

This means that we can obtain the pressure's value if we know the above quantities. Since $R = 8, 2 \frac{J}{m o l K}$ $R = 8,2 J/(molK)$ is constant and we know the values for $T$ $T$ and $V$ $V$ , we must figure out what happens to the number of mols of each gas after the reaction takes place.

Since it takes 5 molecules of gas $B$ $B$ to bind to 2 molecules of gas $A$ $A$ , we can extend this proportion to their corresponding number of mol. Since initially we have $n_{1 A} = 8$ $n_(1A) = 8$ mol and $n_{1 B} = 10$ $n_(1B) = 10$ mol, then, after the reaction takes place, we will end up with $n_{2 A} = 4 m o l$ $n_(2A) = 4 mol$ , $n_{2 B} = 0$ $n_(2B) = 0$ mol and $n_{C} = 2$ $n_C = 2$ mol of the new gas, the product of the $A$ $A$ and $B$ $B$ reaction.

Then, our final pressure will be described by:

$P_{2} = \frac{R T}{V} (n_{A} + n_{B} + n_{C})$ $P_2 = (RT)/V(n_A + n_B + n_C)$ .

(Remember that $P_{1} = \frac{R T}{V} (n_{A} + n_{B})$ $P_1 = (RT)/V(n_A + n_B)$ .

Taking the values to perform the calculations:

$P_1 = (0.082 (atm cancel(L))/(cancel(mol)cancel(K)) * 360 cancel(K))/(5 cancel(L)) * (8 + 10) cancel(mol)$ ;

$P_1 = 106.3 atm$ .

After the reaction takes place:

$P_2 = (0.082 (atm cancel(L))/(cancel(mol)cancel(K)) * 270 cancel(K))/(5 cancel(L)) * (4 + 0 + 2) cancel(mol)$ ;

$P_2 = 26.6 atm$ .

Therefore, the change in pressure, $DeltaP$ , is:

$DeltaP = P_2 - P_1$ ;

$DeltaP = -79,7 atm$ .

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A $5 L$ $5 L$ container holds $8$ $8$ mol and $10$ $10$ mol of gasses A and B, respectively. Every five of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from $360^{o} K$ $360^oK$ to $270^{o} K$ $270 ^oK$ . By how much does the pressure change?

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A 5 L5L5 L container holds 8 88 mol and 10 1010 mol of gasses A and B, respectively. Every five of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from 360^oK360oK360^oK to 270 ^oK270oK270 ^oK. By how much does the pressure change?

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A $5 L$ $5 L$ container holds $8$ $8$ mol and $10$ $10$ mol of gasses A and B, respectively. Every five of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from $360^{o} K$ $360^oK$ to $270^{o} K$ $270 ^oK$ . By how much does the pressure change?