You have calculated ##\Delta V\Delta p##. That is not the same as ##\int p.dV##.kent davidge said:
In the graph, p is the y ordinate and V the x ordinate. So the integral is equivalent to ##\int y.dx##. what's a geometric interpretation of that integral?kent davidge said:
Yes, but you first have to turn the graph into an equation relating p to V. Then plug that function into ##\int p.dV##.kent davidge said:
kent davidge said:
Work done by a gas is a measure of the energy transferred by the gas due to its change in volume or pressure. It is a form of mechanical work and is measured in joules (J).
The work done by a gas can be calculated using the formula W = PΔV, where W is the work done, P is the pressure of the gas, and ΔV is the change in volume of the gas. Alternatively, it can also be calculated using the formula W = ∫PdV, where ∫ represents the integral sign and PdV represents the change in pressure and volume over a given process.
Yes, work can be done on a gas. This occurs when external forces act on the gas, causing it to change in volume or pressure. In this case, the work done on the gas would be negative, as the gas is gaining energy from the external forces.
Examples of work done by a gas include a gas expanding against a piston, a gas compressing in a cylinder, or a gas being released from a pressurized container. Examples of work done on a gas include a gas being compressed by a piston, or a gas being pumped into a pressurized container.
The first law of thermodynamics states that energy cannot be created or destroyed, but it can be transferred or converted from one form to another. The work done by or on a gas is a form of energy transfer, and it contributes to the change in internal energy of the gas, as stated by the first law of thermodynamics.