A $5$ $5$ $L$ $L$ container holds $8$ $8$ mol and $5$ $5$ mol of gasses A and B, respectively. Groups of five molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $100$ $100$ $K$ $K$ to $240$ $240$ $K$ $K$ . How much does the pressure change?

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A $5$ $5$ $L$ $L$ container holds $8$ $8$ mol and $5$ $5$ mol of gasses A and B, respectively. Groups of five molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $100$ $100$ $K$ $K$ to $240$ $240$ $K$ $K$ . How much does the pressure change?

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Answer 1 · 2017-12-26T23:07:02

The reaction decreases the number of moles of gas in the (constant volume) container from $13$ $13$ to $7$ $7$ while increasing the temperature from $100$ $100$ $K$ $K$ to $240$ $240$ $K$ $K$ .

The pressure change is $2794 - 2162 = 632$ $2794-2162=632$ $k P a$ $kPa$

Explanation:

First we need to write a balanced chemical equation for the reaction:

$2 A + 5 B \to A_{2} B_{5}$ $2A+5B->A_2B_5$

Since we have $5$ $5$ mol of $B$ $B$ , all of it will be used up in the reaction. Only $2$ $2$ of the $8$ $8$ mol of $A$ $A$ will be used, so $6$ $6$ mol will remain.

After the reaction, therefore, there will be $6$ $6$ mol of $A$ $A$ and $1$ $1$ mol of $B$ $B$ for a total of $7$ $7$ mol. There were initially $13$ $13$ mol of gas in the container.

We remember that a mole of any gas that behaves like an ideal gas takes up the same volume under the same temperature and pressure conditions.

Let's call the pressure before the reaction $P_{1}$ $P_1$ and after the reaction the pressure is $P_{2}$ $P_2$ .

$P V = n R T$ $PV=nRT$ where $R$ $R$ is the gas constant, $8.314$ $8.314$ $L k P a K^{- 1} m o l^{- 1}$ $LkPaK^-1mol^-1$

$P_{1} = \frac{n R T}{V} = \frac{13 \times 8.314 \times 100}{5} = 2162$ $P_1=(nRT)/V=(13xx8.314xx100)/5=2162$ $k P a$ $kPa$

$P_{2} = \frac{n R T}{V} = \frac{7 \times 8.314 \times 240}{5} = 2794$ $P_2=(nRT)/V=(7xx8.314xx240)/5=2794$ $k P a$ $kPa$

The pressure change is $2794 - 2162 = 632$ $2794-2162=632$ $k P a$ $kPa$

Answer 2 · 2017-12-27T00:23:23

29% increase

Explanation:

Idea gas law:

PV=nRT

For mix gases inside the same volume and temperature

PV $= (n_{A} + n_{B}) R T$ $= (n_A+n_B)RT$

Let $(n_{1}, P_{1}, V_{1}, T_{1}) and (n_{2}, P_{2}, V_{2}, T_{1})$ $(n_1, P_1, V_1, T_1) and (n_2, P_2, V_2, T_1)$ denote the number of moles, pressure, volume, and temperature before and after the reaction happened respectively.

Before the reaction, there are

$n_{1} = n_{A} + n_{B} = 8 + 5 = 13 m o l$ $n_1= n_A+n_B= 8 + 5= 13 mol$ of mixed gases

The reaction

$cancel(5B -> 2A)$ (This reaction is incorrect, I misread the question, B binds with A, not binds to form A)

$5B + 2A -> A_2B_5$

Takes away 5 moles of B and 2 moles of A into one moles of $A_2B_5$ , leaving 6 mol of A.

$n_2= n_A+n_(A_2B_5)$ = 6+1 = 7 mol of mixed gases.

The idea gas law before and after the reaction are:

$P_1V_1 = n_1 RT_1$

$P_2V_2 = n_2RT_2$

$(P_2V_2)/(P_1V_1) = n_2/n_1(RT_2)/(RT_1)$

For a fixed container, $V_1 = V_2$

$P_2/(P_1) = 7/13 * (240K)/(100K) = 1.29$

The change in pressure is

$(P_2-P_1)/p_1 = (DeltaP)/P_1 =P_2/P_1 -1 = 0.29 or 29%$

The pressure has thus increased 85%.

Since the volume of the container is given, $P_1$ can be reality calculated to determine the absolute change in pressure ( $DeltaP$ )

$P_1 = (n_1 RT_1)/V = ((13mol*8.3145 J/(mol K)*100K)/(5L*0.001m^3/L)) /(1.01325 "x" 10^5 N/(m^2atm)) ~ 21.34 atm$

$DeltaP ~0.29P_1= 0.29* 21.3 atm = 6.2 atm$

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