Are you sure that you have the question copied correctly. In particular, could you check that the two highlighted quantities should not be volumes?quietrain said:Homework Statement
calculate work done to compress 1mole of ideal gas from 1atm to 0.5atm at 298K
where it is subjected to constant external pressure of 1atm
Homework Equations
Won = -Integral pdV = -(1atm)(Vf-Vi) = -p(RT/pf - RT/pi) = 2270J
where i have assumed T does not change, is it correct to assume so?
because i read that T changes proportionally with V, if so, how do i find what is my (Vf-Vi)
thanks!
That would be my guess. There are two conflicting statements. The first specifies a change in pressure, whilst the second asserts a constant pressure.quietrain said:yes that's what the question wrote
thats why i found it strange.
shouldn't it be vol, so that its just P(vf-vi)?
PV work done refers to the work done by a system as a result of a change in its volume and pressure. It is a measure of the energy transferred to or from the system due to changes in its internal state.
PV work is calculated by multiplying the change in the system's volume by the change in its pressure. In equation form, it can be written as W = PΔV, where W represents work, P represents pressure, and ΔV represents change in volume.
The units of PV work are joules (J) in the SI system. In some cases, it may also be expressed in other units such as kilojoules (kJ) or calories (cal).
PV work is a key concept in thermodynamics, as it represents the energy transferred in a system due to changes in its volume and pressure. It is used to analyze and understand processes such as compression, expansion, and heat transfer in various systems.
Yes, PV work can be negative. This occurs when the system does work on its surroundings, rather than the other way around. This may happen, for example, when a gas expands against an external pressure. In this case, the work done by the gas is negative, as it is losing energy to the surroundings.