Thursday, October 31, 2013
Our third law that we will concern ourselves with is the first law of thermodynamics. It will be the third of the “big six” we will look at for now. Here’s the law.
This law says that the heat added to a certain mass of a gas is equal to its change in internal energy + the work done BY the gas ON the environment.
A classic example of the first law of thermodynamics at work is the piston. In order to make the piston system move, we first need to add heat to it. All or some of this heat Q is used to increase the internal energy of m. The rest of this heat is used to move the piston upwards.
In order to move a piston up, we need to do work on it. To push the piston outward through an external force dx, we set up the following equation, where dW equals the derivative of work and dx equals the derivative of x. For those of you who don’t know calculus, derivative means “rate of change.”
dW = Fdx
In the case of the piston, we have a cross-sectional area A exposed, so because F=p*A, we plug that in to get dW=p*A*dx = p*dV. Our final equation, dW = p*dV, says that the work done by the substance when its volume increases by a small increment dV is equal to the pressure of the substance multiplied by its increase in volume. If you want to find the total work done, just take the integral from V_1 to V_2.
I thought that was pretty easy to understand. Then we started getting into some other proofs that looked quite ugly.
Just to review:
Work = F*d (force*distance) = ρ*ΔV (density*change in volume)
α = V/m (volume/mass)
Below are some variations on the ideal gas law. C_v is equal to the specific heat of vapor (water vapor), which is the amount of heat per unit mass required to raise the temperature by one degree Celsius. dq is the change in thermal energy, dT is the change in temperature, dα is the change in specific volume (1/density), and dp is the change in pressure.