An ideal gas is in contact with a heat reservoir so that it remains at a constant temperature of 100 K. The gas is compressed from a volume of 34 L to a volume of 17 L. During the process the mechanical device pushing the piston to compress the gas is found to expend 3 kJ of energy. What is the magnitude of the heat flow between the heat reservoir and the gas and in what direction does the heat flow occur?I honestly have no idea how to solve it. Please help!
Apply the first law of thermodynamics equation for heat (Q), work (W), and internal energy (U). Q = deltaU + W In an ideal gas, U is exclusively a function of temperature. Because the process is isothermal, U is constant. Therefore deltaU = 0. Note: this is only true if the process is quasistatic and frictionless, which means slow enough that the pressure can fully distribute at every point along the way. And our first law equation becomes: Q = W We are given the mechanical work done on the gas. Recall the sign convention for positive energy: Positive heat is heat added to the system Positive work is work done by the system Therefore W = -3 kJ Since Q = W Q = -3 kJ Result: 3 kJ of heat are transferred. Interpreting the negative sign this means that 3 kJ escape the system of gas and transfer to the thermal reservoir.
Google pv=nrt, and it has online calculators and links. YOur question is a little different, becasue we are effectively ignoring the T, as the heat generated by compression is drawn off by the heat reservoir. Normally, the gas would heat up as you compressed it. But the heat would flow out to the reservoir to maintain 100 K. I don't know in this case how much of the 3 kJ is wasted to the reservoir, and how much remains in the potential energy of the pressurized gas. It depends on how fast the plunger is being pushed. If infinitely slow, then all 3kJ is preserved, and none goes to the heat exchanger. If instantaneous, it all goes to heat and is lost.
