A #4 L# container holds #4 # mol and #4 # 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 #280^oK# to #320^oK#. By how much does the pressure change?

Question

A #4 L# container holds #4 # mol and #4 # 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 #280^oK# to #320^oK#. By how much does the pressure change?

Olivia Askew · Accepted Answer

#10 atm# First, you need to turn the information into a chemical equation: #3A(g) + 4B(g) -> A_3B_4(g)# Then you find the moles of product formed and moles of reactant left. You should get 1 mol of #A_3B_4(g)# and 1 mol #A(g)# left. After that you just input the information into the formula #P_(Total) = (n_(Total)RT)/V# #P_(Total) = (2mols * 0.08206 "L atm " mol^-1 K^-1 * 320 K)/(4L)# #P_(Total) = 13.1296atm# Since you only have one significant figure, the answer would be #10 atm#.

Adam Arnold · Answer

undefined To solve this problem, we can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin.

First, we calculate the initial pressure (P1) and final pressure (P2) using the initial and final temperatures, respectively.

Given:
Initial temperature (T1) = 280 K
Final temperature (T2) = 320 K

Using the ideal gas law, we can find the initial pressure (P1) and final pressure (P2):

P1 = (n1 + n2)RT1 / V
P2 = (n1 + n2)RT2 / V

Where:
n1 = 4 mol (moles of gas A)
n2 = 4 mol (moles of gas B)
R = gas constant = 0.0821 atm L / (mol K) (assuming pressure is in atm and volume is in liters)

Now, we plug in the values:

P1 = (4 + 4) * 0.0821 * 280 / 4 = 45.684 atm
P2 = (4 + 4) * 0.0821 * 320 / 4 = 52.236 atm

The pressure change (ΔP) can be calculated as:
ΔP = P2 - P1

ΔP = 52.236 atm - 45.684 atm = 6.552 atm

Therefore, the pressure changes by approximately 6.552 atm.