Sample of gas is compressed at a constant temperature

[Miike012]

Problem: When a sample of gas is compressed at a constant temperature, the product of the pressure and the volume remains constant:

PV = C

Q: A sample of gas is in a container at low pressure and its steadily compressed at constant temperature for 10 minutes. Is the volume decreasing more rapidly at the beginning of 10 minutes or at the end of 10 minutes?

A: dV/dt = -C/P^2(dP/dt)

By looking at my graph... at the end of 10 seconds dV/dt is approaching smaller and smaller values... thus I picked "end of 10 seconds" ... but the answer is beginning of 10 seconds... what did I do wrong?

 

[verty]

Hmm, the question says "steadily compressed" which I take to mean that dP/dt is a constant function. But... (removed).

Aargh, sorry, it's late, ignore this.

 

[Mark44]

Problem: When a sample of gas is compressed at a constant temperature, the product of the pressure and the volume remains constant:

PV = C

Q: A sample of gas is in a container at low pressure and its steadily compressed at constant temperature for 10 minutes. Is the volume decreasing more rapidly at the beginning of 10 minutes or at the end of 10 minutes?

A: dV/dt = -C/P^2(dP/dt)

By looking at my graph... at the end of 10 seconds dV/dt is approaching smaller and smaller values... thus I picked "end of 10 seconds" ... but the answer is beginning of 10 seconds... what did I do wrong?

I'm assuming this is a graph of dV/dt versus t.

At the beginning of the 10-minute interval, dV/dt is smaller (more negative) than at the end of the interval. This means that V is decreasing more at the beginning of the interval than at the end.

There is a connection between a the graph of a function and its derivative.

f' > 0 on an interval ==> f is increasing on that interval
f' < 0 on an interval ==> f is decreasing on that interval

 

[HallsofIvy]

From PV= C, the pressure increases as the volume decreases. At the end of the compression, P will be greater than at the beginning so that 1/P^2 will be less.