A #5 L# container holds #15 # mol and #6 # mol of gasses A and B, respectively. Every three of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from #360K# to #210K#. By how much does the pressure change?

2 Answers
Apr 5, 2016

Answer:

The pressure changed by #8.03 × 10^6 Pa#.

Explanation:

To get started we’ll work out how many moles of gas are present at the end of the reaction. There are 6 moles of gas B. In the reaction gas B bonds with gas A at a ratio of 3:2 molecules, so that ratio will apply to the number of moles also. 3 moles of gas B will bond with 2 moles of gas A. In total all 6 moles of gas B will bond with 4 moles of gas A and the product of that reaction will be 2 moles of a new gas. Let’s call the new gas, gas C.

That leaves us with 15 – 4 = 11 moles of gas A, 0 moles of gas B (it all reacted), and 2 moles of gas C. So the final number of moles of gas will be 13 moles. The initial number for moles was 15 + 6 = 21 moles.

Calculate the Final Pressure
To solve the problem of the pressure change we will use the ideal gas equation:
#(pV)/(nT) =# constant.
You may be familiar with the equation stated as: #pV = nRT# where R is the molar gas constant.

Since the LHS of the equation is equal to a constant we can now state the equation like this:
#(p_1V_1)/(n_1T_1) = (p_2V_2)/(n_2T_2)#

The volume does not change so #V_1=V_2# and they will cancel out of the equation. We need to find the final pressure so rearrange for #p_2#:
#p_2 = p_1(V_1n_2T_2)/(V_2n_1T_1) = p_1(n_2T_2)/(n_1T_1)#

The above equation will give us p₂ as a multiple of p₁. Now substitute the values into the equation:
#p_2 = p_1(13 × 210)/(21 × 360) = 0.3611 p_1#

Pressure Change
To calculate by how much the pressure changes find the difference between initial and final pressures:
#p_1 – p_2 = p_1 – 0.3611 p_1 = 0.639 p_1#

Calculate the value of p₁ by using the ideal gas equation:
Volume conversion: # V = 5 L = 5 × 10^(-3) m³#

#p_1V_1 = n_1RT_1 ⇒ p_1 = (n_1RT_1) / V_1 = (21 × 8.31 × 360) / (5 × 10^(-3)) = 12.565 × 10^6 Pa#

The pressure therefore changed by:
#0.639 × 12.565 × 10^6 = 8.03 × 10^6 Pa#.

enter image source here
As explained in the above figure
Initial number of moles of gas molecules was #n_1=21 #moles
and Final number of moles of gas molecules was #n_2=13# moles

We Know from Equation of state for ideal gas

#PV = nRT#,

where:

  • P = pressure in atm
  • V = volume in L
  • T = temperature in K
  • n = number of moles
  • R = universal gas constant.=0.082#LatmK^-1mol^-1#

For Initial state
If pressure is #P_1#
Volume #V =5L#
Number of moles #n_1=21 #
Temperature #T_1=360K#
and then #P_1=(n_1RT_1)/V#

For Final state
If pressure is #P_2#
Volume #V =5L#
Number of moles #n_2=13 #
Temperature #T_2=210K#
and then #P_2=(n_2RT_2)/V#

So decrease in Pressure:

#= P_1-P_2=R/Vxx(n_1T_1-n_2T_2)#
#= 0.082/5xx(21xx360-13xx210)# #atm#
#= 0.082/5xx4830# #atm=79.212# #atm#
#= 79.212# #atm#

When we convert this to #Pa#, we get:

#79.212# #cancel(atm)xx101.325xx10^3# #(Pa)/cancel(atm)#

#~~ 8026xx10^3# #Pa#