A #6 L# container holds #4 # mol and #6 # mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from #480^oK# to #270^oK#. How much does the pressure change?

Question

A #6 L# container holds #4 # mol and #6 # mol of gasses A and B, respectively.  Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from #480^oK# to #270^oK#. How much does the pressure change?

William Audy · Accepted Answer

Final Pressure : #P_f = (\frac{n_fT_f}{n_iT_i})P_i = 0.3765# #MPa#
Pressure Change :
#\DeltaP = P_f-P_i = [\frac{n_fT_f}{n_iT_i}-1]P_i = -2.9706# #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: #4A+6B ightarrow 2A_2B_3# #4# mols of #A# combine with #6# mols of #B# to give #2# mols of #A_2B_3# Before Reaction: #P_iV=(n_a+n_b)RT_i# #n_a = 4# #mols; \quad n_b = 6# #mols; \qquad n_i=n_a+n_b=10# #mols# # T_i=480 K; \qquad V=6L=6 imes10^{-3}m^3# #P_i = (n_a+n_b)\frac{RT_i}{V} = 3.3472 imes10^6# #Pa = 3.3472# #MPa#. After Reaction: #P_fV = n_{ab}RT_f# #n_f = n_{ab} = 2# #mols;# # T_f=270 K; \qquad V=6L=6 imes10^{-3}m^3# #P_f = (\frac{n_fT_f}{n_iT_i})P_i = (\frac{2 imes270K}{10 imes480K}) imes3.3472 # #MPa# #\qquad= 0.3765# #MPa# # \DeltaP = P_f-P_i = [\frac{n_fT_f}{n_iT_i}-1]P_i = -2.9706# #MPa# The fact that the pressure change is negative means that the pressure is dropping.

Charles Aeschliman · Answer

undefined To find the change in pressure, we can use the ideal gas law, which states:

$$ PV = nRT $$

Where:
- $ P $ is the pressure,
- $ V $ is the volume,
- $ n $ is the number of moles,
- $ R $ is the gas constant,
- $ T $ is the temperature.

Since the volume, number of moles, and gas constant remain constant in this case, we can use the relationship between pressure and temperature directly:

$$ \frac{P_2}{P_1} = \frac{T_2}{T_1} $$

Given that the initial temperature $ T_1 = 480 \, \text{K} $ and the final temperature $ T_2 = 270 \, \text{K} $, we can calculate the change in pressure using:

$$ \frac{P_2}{P_1} = \frac{270 \, \text{K}}{480 \, \text{K}} $$

$$ P_2 = P_1 \times \frac{270 \, \text{K}}{480 \, \text{K}} $$

Now, we need to find the initial pressure ($ P_1 $). Since we have 4 moles of gas A and 6 moles of gas B in a 6 L container, the total number of moles is 10. Using the ideal gas law with the initial temperature ($ T_1 = 480 \, \text{K} $) and the number of moles ($ n = 10 $):

$$ P_1 = \frac{nRT}{V} = \frac{(10 \, \text{mol}) \times (8.314 \, \text{J/mol·K}) \times (480 \, \text{K})}{6 \, \text{L}} $$

Now, we can calculate $ P_2 $ using the formula we derived earlier:

$$ P_2 = P_1 \times \frac{270 \, \text{K}}{480 \, \text{K}} $$

$$ P_2 = \left( \frac{(10 \, \text{mol}) \times (8.314 \, \text{J/mol·K}) \times (480 \, \text{K})}{6 \, \text{L}} \right) \times \frac{270 \, \text{K}}{480 \, \text{K}} $$

After calculating $ P_1 $ and $ P_2 $, we can find the change in pressure ($ \Delta P $):

$$ \Delta P = P_2 - P_1 $$

Substitute the values to calculate the change in pressure.

Charlotte Aaron · Answer

undefined According to the ideal gas law, $ PV = nRT $, where $ P $ is pressure, $ V $ is volume, $ n $ is the number of moles, $ R $ is the gas constant, and $ T $ is temperature.

Since the volume and the number of moles of gas A and B remain constant, we can use the formula $ P_1/T_1 = P_2/T_2 $ to find the pressure change, where $ P_1 $ and $ T_1 $ are the initial pressure and temperature, and $ P_2 $ and $ T_2 $ are the final pressure and temperature, respectively.

Given that the initial temperature $ T_1 = 480 $ K and the final temperature $ T_2 = 270 $ K, we can use the formula to calculate the pressure change.