A 5 L5L container holds 15 15 mol and 6 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 360K360K to 210K210K. By how much does the pressure change?

2 Answers
Apr 5, 2016

The pressure changed by 8.03 × 10^6 Pa8.03×106Pa.

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) =pVnT= constant.
You may be familiar with the equation stated as: pV = nRTpV=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)p1V1n1T1=p2V2n2T2

The volume does not change so V_1=V_2V1=V2 and they will cancel out of the equation. We need to find the final pressure so rearrange for p_2p2:
p_2 = p_1(V_1n_2T_2)/(V_2n_1T_1) = p_1(n_2T_2)/(n_1T_1)p2=p1V1n2T2V2n1T1=p1n2T2n1T1

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_1p2=p113×21021×360=0.3611p1

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.=0.082LatmK^-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