A #9 L# container holds #8 # mol and #2 # mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from #370^oK# to #425 ^oK#. How much does the pressure change by?

Question

A #9 L# container holds #8 # mol and #2 # mol of gasses A and B, respectively.  Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from #370^oK# to #425  ^oK#. How much does the pressure change by?

Leo Abeyta · Accepted Answer

#P_i = (n_a+n_b)\frac{RT_i}{V} = 1.7201# #MPa#.
#P_f = (n_{ab}+n_c)\frac{RT_f}{V} = 1.5806# #MPa#

# \DeltaP = P_f-P_i = -0.1395# #MPa# 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: 8 moles of #A# and 2 moles of #B# combine to give 1 mole of #AB_2# and 7 moles of #A#. #8A+2B ightarrow 7A + AB_2# Before Reaction: #P_iV=(n_a+n_b)RT_i# #n_a = 8# #mols; \quad n_b = 2# #mols; \qquad n_i=n_a+n_b=10# #mols# # T_i=370 K; \qquad V= 9L = 9 imes10^{-3}m^3# #P_i = (n_a+n_b)\frac{RT_i}{V} = 1.7201 imes10^6# #Pa = 1.7201# #MPa#. After Reaction: #P_fV = n_f RT_f# #n_{ab} = 1# #mols; \quad n_c = 7# #mols; \qquad n_f = n_{ab} + n_c = 8# #mols# # T_f=425 K; \qquad V = 9L = 9 imes10^{-3}m^3# Because #R# and #V# are constants, #P_f = (\frac{n_fT_f}{n_iT_i})P_i = (\frac{8 imes425K}{10 imes370K}) imes1.7201 # #MPa# #\qquad= 1.5806# #MPa# # \DeltaP = P_f-P_i = [\frac{n_fT_f}{n_iT_i}-1]P_i = -0.1395# #MPa# The pressure change is negative, indicating that the pressure decreases.

William Audy · Answer

undefined The pressure increases by approximately $ 6.8 \, \text{kPa} $.