- #1
Sudikshya Pant
- 14
- 0
A quantity of ideal gas undergoes an expansion that doubles its volume. Does the gas do more work on its surroundings if the expansion is at constant pressure or at constant temperature?
The answer in the book says W at constant temperature gives a greater value for the given case.
But when I tried to solve it mathematically, I got greater value of W at constant pressure which is plain wrong. I don't understand where did I go wrong.
For constant pressure my result was:
W = p(2V-V) =pV = (nRT/V)*V =nRT
For constant temperature my result was:
W= nRT ln(2V/V) =nRT ln(2) =0.69 nRT
So, I got lesser value for constant temperature than constant pressure.
Where did I went wrong?
The answer in the book says W at constant temperature gives a greater value for the given case.
But when I tried to solve it mathematically, I got greater value of W at constant pressure which is plain wrong. I don't understand where did I go wrong.
For constant pressure my result was:
W = p(2V-V) =pV = (nRT/V)*V =nRT
For constant temperature my result was:
W= nRT ln(2V/V) =nRT ln(2) =0.69 nRT
So, I got lesser value for constant temperature than constant pressure.
Where did I went wrong?