1. Change in volume is essentially due to formation of gas
QUESTION:
Estimate the change in energy due to PV work for a reaction in aqueous solution that generates 0.50 moles of gas at ordinary conditions (298K, atmospheric pressure) A kJ, B kJ, C kJ, D. –5.0 kJ
w = - Pex dV = - Pex V
Change in volume is essentially due to formation of gas
V = volume of gas generated = ngas RT/P
Pressure of gas = external pressure: P = Pex
w = - P (ngas RT/P) = -ngas RT
Estimate the change in energy due to PV work for a reaction in aqueous solution that generates 0.50 moles of gas at ordinary conditions (298K, atmospheric pressure)
PAUSE
Work due to expansion or compression of an object is called “pressure volume work”,… or PV work
Using modern sign convention, PV work…
CLICK is calculated by taking the negative of the integral of the external pressure with respect to volume.
HIGHLIGHT integral
If the external pressure is constant, which is typically the case for reactions carried out in containers that are open to the atmosphere,
Then the formula simplifies to
HIGHLIGHT Pex V
The negative of the external pressure times the change in volume.
The external pressure is equal to the atmospheric pressure.
CLICK
Here’s an illustration of the reaction that generates a gas at constant pressure. Imagine the reaction mixture
as being in a container with a movable, frictionless piston, shown here in red.
HIGHLIGHT piston
As the reaction generates a gas, it pushes the piston upward.
If the pressure on the piston is constant and equal to the atmospheric pressure, then the work done by the system
is essentially the same as the case where the container is open to the atmosphere.
At this point, we should be able to eliminate two of the answer choices given.
The production of gas leads to an increase in volume…. Therefore, delta V must be positive.
CALLOUT “+” pointing to V
Since external pressure is always a positive number…
CALLOUT “+” pointing to Pex
the calculated work must be negative.
Whenever a system expands against an external pressure, the system is doing work,
which involves a loss of energy. By convention, work is assigned a negative value
Therefore, choices B and C are incorrect. Both of these are positive. We expect a negative value for w.
CLICK Since the change in volume is essentially due to to the formation of the gaseous product,
We can say that
CLICK The change in volume … deltaV HIGHLIGHT
is equal to the volume of the gas produced.
Using the ideal gas equation, PV equals nRT, we can calculate the volume of the gas as…
moles of gas….
HIGHLIGHT ngas
times RT over P.
where R is the ideal gas constant,
CALLOUT BOX pointing to R “R = L atm mol-1 K-1, R = L bar mol-1 K-1,
R = J mol-1 K-1”
T is the temperature of the gas in Kelvin, CALLOUT “298K” pointing to T
and P is the pressure of the gas.
CALLOUT “Pressure” pointing to P
The pressure of the gas is equal to the external pressure.
HIGHLIGHT P=Pex
CLICK Therefore, we can substitute P for external pressure….
HIGHLIGHT P in w=-P (ngas RT/P) and P in –PexV
and substitute the volume of the gas for the change in volume
HIGHLIGHT ngas RT/P in w=-P (ngas RT/P) and V in –PexV
Since P cancels out…
SHOW CANCELLATION of P in -P (ngas RT/P),
we can calculate work as the negative of
moles of gas times R times T
HIGHLIGHT –ngas RT
It’s convenient to memorize the value of RT At 298 Kelvin. It’s equal to 2.48 kilojoules per mole.
Therefore, work is equal to the negative of …
the moles of gas produced… 0.50 moles
HIGHLIGHT times the value of RT at 298 Kelvin…. Which is 2.48 kilojoules per mole.
HIGHLIGHT
This gives us an answer of 1.2 kilojoules.
The correct answer is A.
CLICK PAUSE END RECORDING
At K: RT = 2.48 kJ/mol
w = -(0.50 mol) (2.48 kJ/mol)
= kJ

2. Video ID: © 2008, Project VALUE (Video Assessment Library for Undergraduate Education), Department of Physical Sciences Nicholls State University Author: Glenn V. Lo Assisted by Johuan Gasery In partial fulfillment of the requirements for CHEM 481, Fall 2008.