Question
Please show all work and detailed explanation I will rate.
In the classic TV series Batman, story lines almost always lasted for two episodes. At the end of the first episode, Batman and Robin were invariably caught in some diabolical trap intended to end their lives. In one episode, Egghead decided to both roast and crush the dynamic duo. He placed the caped crusaders in a room whose temperature steadily rose as two of the walls closed in on the men in tights. According to the ideal gas law, the pressure (P), temperature (T), and volume (V) inside the room were related by the equation PV/T = k where k is a constant specific to the types and amounts of gas that were in the room. When the heroes were placed in the room, the room was 2m times 2m times 2m, the temperature in the room was 20degree C, and the air pressure in the room was 100 kPa. Once Egghead activated the trap, two of the walls moved toward each other at the constant rate of .1 m/min (in total) and the temperature in the room rose at a constant rate of 2degree C/min. Since the volume and temperature both changed at constant rates, it might seem intuitive to you that the pressure also changed at a constant rate. Let's see if that is in fact the case. Use the initial conditions in the room to determine the value of k (without unit) and rewrite Use the initial conditions in the room to determine the value of k (without unit) and rewrite the ideal gas law using this value for k. Explicitly define P, T, and V (including units) in terms of the amount of time that had passed since the trap was set in motion. Your rate equation comes from differentiating the ideal gas law equation with respect to time. There is a simple algebraic adjustment you can make to the equation that will greatly simplify this task. Go ahead and adjust the equation and then find your rate equation. Determine the values of P, T, and V at both five minutes and eight minutes into the motion of the trap. State the known rate vales (remembering to think about the sign on each rate). You might want to think twice about the value of dV/dt. Use all of these values along with the rate equation to determine the value of dP/dt at both five minutes and eight minutes into the motion of the trap.
Explanation / Answer
1. k= 100kPa * 8/(293 K) =2730.3
3. T = 20+2t, V = (2-.2t)^3, P = kT/V
2. PV=kT => VdP/dt + P dV/dt = kdT/dt
one can get P(t) from here
4. after 5 min : T = 30, V = 1, P = 81909
similarly cab be calculated for 8 min
