How much work is done by the system when
2.00 moles of O2 expand from a volume of
3.00 liters to a volume of 9.6 liters against a constant external pressure of 0.0100 atm
answer in units of joules
7 years late but i have the same question as alex
This may be a little late but what about moles
w = -pdV
Work done by the system is -
pdV is 0.01*(9.6-6.6)
where did you get the 6.6 from? and i did the work correctly and the answer was wrong.
I typed in the wrong number. That's V2-V1 = (9.6-3.00) so
work = -0.01*6.6 = ? in L*atm since 0.01 is in atm and delta V is L. The problem asks for joules. Convert L*atm to Joules.
1 L*atm = 101.325 joules.
To determine the work done by the system, we can use the formula:
Work = -Pext * ΔV
where:
- Work is the work done by the system
- Pext is the external pressure
- ΔV is the change in volume
In this case, the external pressure is given as 0.0100 atm. However, it is important to note that the units of work should be in joules. Since 1 atm = 101325 Pa and 1 J = 1 kg m²/s², we need to convert atm to Pa to ensure consistent units.
1 atm = 101325 Pa
So, the external pressure of 0.0100 atm can be converted to:
Pext = 0.0100 atm * 101325 Pa/atm
Pext = 1013.25 Pa
Now, we can calculate the change in volume (ΔV) by subtracting the initial volume (V1) from the final volume (V2):
ΔV = V2 - V1
ΔV = 9.6 L - 3.0 L
ΔV = 6.6 L
It is important to note that the volume must be in a unit consistent with the pressure, so we will convert liters (L) to cubic meters (m³):
ΔV = 6.6 L * (1 m³ / 1000 L)
ΔV = 0.0066 m³
Finally, we can calculate the work done by the system:
Work = -Pext * ΔV
Work = -(1013.25 Pa) * (0.0066 m³)
Now, we can multiply these two values to find the work done:
Work = -6.6669 J
Since work is a scalar quantity, we disregard the negative sign and express the value as:
Work = 6.6669 J
Therefore, the work done by the system when 2.00 moles of O2 expand from a volume of 3.00 liters to a volume of 9.6 liters against a constant external pressure of 0.0100 atm is approximately 6.6669 joules.