A #4 L# container holds #19 # mol and #20 # mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from #210^oK# to #120^oK#. How much does the pressure change?

Question

A #4 L# container holds #19 # mol and #20 # mol of gasses A and B, respectively.  Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from #210^oK# to #120^oK#. How much does the pressure change?

Sebastian Acosta · Accepted Answer

The pressure drops from 168 atm to 22.2 atm.

I'll respond to a query from another Wisconsinite! Determining the change in the number of moles from the reaction is the most challenging aspect of this problem. Please #n_0^A# = the initial number of moles of gas A = 19. #n_0^B# = the initial number of moles of gas B = 20. #n_0# = the total number of moles of gas present before the reaction = 19 + 20 = 39. If we assume that gasses A and B are ideal, we have all the information we need to calculate the initial pressure. (In reality, this is not the case at 120K!) #P^0=(n^0RT^0)/V^0=(39(0.082057)(210))/4~~168# atm The problem indicates that the reaction proceeds as follows: The two types of gases react to form a new gas C. #3A+4BrarrC# Since we only have roughly 5% more B than A prior to the reaction, we know that gas B is the limiting reagent in this case. We will assume that the reaction proceeds to completion in order to eliminate gas B after the reaction. Please #x# = the number of moles of gas B that reacts = 20. #n^A# = the number of moles of A left after the reaction. #n^B# = the number of moles of B left after the reaction = 0. #n^C# = the number of moles of C formed by the reaction. Due to the stoichiometry of the reaction #n^A=n_0^A-(3/4)x=19-(3/4)20=4# moles of gas A. #n^C=x/4=20/4=5# moles of gas C The total number of moles of gas after the reaction, #n#, is #n=n^A+n^B+n^C=4+0+5=9# moles of gas. We can compute the pressure following the reaction, assuming once more that the ideal gas law is true. #P=(nRT)/V=(9(0.082057)(120))/4~~22.2# atm

Adrian Ayala · Answer

undefined To calculate the pressure change, we can use the ideal gas law, $PV = nRT$.

1. First, we need to find the initial and final pressures. 
2. Then, we can find the difference to determine the pressure change.

Given:
- Initial temperature ($T_i$) = 210 K
- Final temperature ($T_f$) = 120 K
- Initial moles of gas A ($n_A$) = 19 mol
- Initial moles of gas B ($n_B$) = 20 mol
- Initial volume ($V$) = 4 L

Using the ideal gas law:
$$P_i = \frac{n_A RT_i}{V} + \frac{n_B RT_i}{V}$$
$$P_f = \frac{n_A RT_f}{V} + \frac{n_B RT_f}{V}$$

$$P_i = \frac{19 \times R \times 210}{4} + \frac{20 \times R \times 210}{4}$$
$$P_f = \frac{19 \times R \times 120}{4} + \frac{20 \times R \times 120}{4}$$

Calculate $P_i$ and $P_f$.

Then, find the pressure change:
$$∆P = P_f - P_i$$