Conditions when qp=qv
Answers
Answered by
7
v) is heat at constant volume and q(p) is heat at constant pressure.
Think about the formula ΔU = q+w. ΔU is the change in internal energy of system. If you think back to the piston model mentioned in lecture, energy used as work of expansion (the gas inside piston is pushing against the piston, causing a positive change in volume, and expanding... doing work from the system on the surroundings) you will find that work = -PΔV. Work is negative because the system loses energy trying to expand. That being said, you can substitute work in the first formula and you will find that ΔU = q-PΔV.
If the change in volume is equal to zero (same thing as a CONSTANT volume) then PΔV will go to zero, making ΔU = q(v), denoting the subscript "v". This is important to distinguish from q(p) because constant volume/pressure would provide different systems. At constant volume in a 'bomb' calorimeter we would expect V to equal 0, so any heat gained or lost is only equal to the change in internal energy.
If the pressure is at CONSTANT pressure, making ΔU = q(p)-PΔV, denoting the subscript "p", then ΔU is also equal to ΔH - PΔV, given by the formula that q(p) is also equal to a change in enthalpy. In this case, PΔV doesn't go to zero because it's not a change in pressure, but we have constant value for pressure so in this equation as constant pressure it accounts for BOTH internal energy and the pressure of a system.
Two systems at rest could have the same internal energy, but be at different pressures, and, thus, have different capacity to do work. Hope I answered your question!
Think about the formula ΔU = q+w. ΔU is the change in internal energy of system. If you think back to the piston model mentioned in lecture, energy used as work of expansion (the gas inside piston is pushing against the piston, causing a positive change in volume, and expanding... doing work from the system on the surroundings) you will find that work = -PΔV. Work is negative because the system loses energy trying to expand. That being said, you can substitute work in the first formula and you will find that ΔU = q-PΔV.
If the change in volume is equal to zero (same thing as a CONSTANT volume) then PΔV will go to zero, making ΔU = q(v), denoting the subscript "v". This is important to distinguish from q(p) because constant volume/pressure would provide different systems. At constant volume in a 'bomb' calorimeter we would expect V to equal 0, so any heat gained or lost is only equal to the change in internal energy.
If the pressure is at CONSTANT pressure, making ΔU = q(p)-PΔV, denoting the subscript "p", then ΔU is also equal to ΔH - PΔV, given by the formula that q(p) is also equal to a change in enthalpy. In this case, PΔV doesn't go to zero because it's not a change in pressure, but we have constant value for pressure so in this equation as constant pressure it accounts for BOTH internal energy and the pressure of a system.
Two systems at rest could have the same internal energy, but be at different pressures, and, thus, have different capacity to do work. Hope I answered your question!
Answered by
2
Answer: when delta value is zero
Explanation . qp= qv+delta nOT
When delta n=0
Then qp=qv
Similar questions