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 5.8 liters against a constant external pressure of 0.0100 atm? answer in units of J
I think this is a duplicate post.
To find the work done by the system, we can use the formula:
Work = -PΔV
where P is the constant external pressure and ΔV is the change in volume.
First, we need to calculate the change in volume:
ΔV = Vf - Vi
= 5.8 L - 3.0 L
= 2.8 L
Substituting the known values into the formula:
Work = - (0.0100 atm)(2.8 L)
Now, we need to convert the units of pressure and volume to their corresponding SI units:
1 atm = 101.325 J/L
1 L = 0.001 m³
Converting:
Work = - (0.0100 atm)(2.8 L)
= - (0.0100 atm)(2.8 L) (101.325 J/L)(0.001 m³/L)
= - (0.0100)(2.8)(101.325)(0.001) J
≈ - 0.00028326 J
Note that the negative sign indicates that work is done on the system, as it is expanding.
Therefore, the work done by the system is approximately -0.00028326 J.