A #7 L# container holds #12 # mol and #18 # 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 #350^oK# to #175 ^oK#. By how much does the pressure change?

Question

A #7 L# container holds #12 # mol and #18 # 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 #350^oK# to #175  ^oK#. By how much does the pressure change?

Elena Albarran · Accepted Answer

#\Delta P = P-P_0 = (2.494-12.47)\quad MPa = -9.976\quadMPa# Ideal Gas Equation of State: #PV=nRT# #R = 8.314\quadJK^{-1}mol^{-1}# Before Reaction: #P_{A0}# - Partial pressure of species A; #P_{B0}# - Partial pressure of species B; #P_0 = P_{A0}+P_{B0}# - Total pressure before reaction; #n_{A0} = 12\quad mols# - Number of moles of species A; #n_{B0}=18\quad mols# - Number of moles of species B; #T_0=350\quad K# - Temperature of the mixture; #V_0=7\quad Lit = 7 imes10^{-3} \quad m^3# - Volume of the mixture; #P_{A0}=(n_{A0}/V)RT_0 = 4.988\quad MPa;\quad # // #1 \quadMPa=10^6\quad Pa# #P_{B0}=(n_{B0}/V)RT_0 = 7.483\quad MPa#; #P_0 = P_{A0}+P_{B0} = 12.47\quad MPa# After Reaction: #6\quad mols# of A would combine with #18\quad mols# of B to form #6\quad mols# of AB. The final mixture would be #6\quad mols# of A and #6\quad mols# of AB. #P_{A}# - Partial pressure of species A; #P_{AB}# - Partial pressure of the new species AB; #P=P_{A}+P_{AB}# - Total pressure after reaction; #n_{A} = 6\quad mols# - Number of moles of species A; #n_{AB}=6\quad mols# - Number of moles of the new species AB; #T=175\quad K# - Temperature of the mixture after reaction; #V=7\quad Lit = 7 imes10^{-3} \quad m^3# - Volume of mixture; #P_{A}=(n_{A}/V)RT = 1.247 \quad MPa#; #P_{AB}=(n_{AB}/V)RT = 1.247\quad MPa#; #P = P_{A}+P_{AB} = 2.494\quad MPa# Change in Pressure : #\Delta P = P-P_0 = (2.494-12.47)\quad MPa = -9.976\quadMPa# So the pressure decreases by 9.976 MPa which in terms of percentage is a decrease by #80%#

Benjamin Asiedu · Answer

undefined The pressure change in this scenario can be calculated using the ideal gas law and the concept of stoichiometry for the reaction between gases A and B. Since three molecules of gas B bind to one molecule of gas A, the total number of gas molecules decreases after the reaction. The ideal gas law equation relating pressure, volume, moles, and temperature is:

$$ PV = nRT $$

Where:
- $ P $ is pressure
- $ V $ is volume (7 L)
- $ n $ is the number of moles
- $ R $ is the ideal gas constant
- $ T $ is temperature (in Kelvin)

The reaction will reduce the number of gas molecules, affecting the total number of moles in the container. Therefore, the pressure change can be calculated using the ideal gas law before and after the reaction, considering the temperature change as well. The ideal gas law can be rearranged to solve for pressure:

$$ P = \frac{nRT}{V} $$

Given that $ n_A = 12 $ mol and $ n_B = 18 $ mol initially, and the temperature changes from 350 K to 175 K, we can calculate the initial and final pressures using the ideal gas law. Then, the pressure change can be determined.

$$ P_1 = \frac{(12 + 18) \times R \times 350}{7} $$

$$ P_2 = \frac{(12 + 18) \times R \times 175}{7} $$

$$ \Delta P = P_2 - P_1 $$

Where $ \Delta P $ is the pressure change.