# What is the final volume of the container?

## Homework Statement

A container has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction and is placed below a heater.
Suppose the heater has 25 W of power and are turned on for 15 s. What is the final volume of the container?

Initial Volume: 800 cm$$^{3}$$
Initial Pressure: 1.11*10$$^{5}$$ Pa
Initial Temperature: 20 C

## Homework Equations

pV = nRT
Q = p_{power}$$\Delta$$t = nC_{p}$$\Delta$$T

## The Attempt at a Solution

Using pV = nRT and knowing that p_i = p_f I can get V_{f} = V_{i}T_{f} / T_{i}

I now need T_{f}

To get that I used
p_{power}$$\Delta$$t = nC_{p}$$\Delta$$T

to get

T_{f}] = p_{power}$$\Delta$$t/nC_{p}] + T]_{i}

Then I used p_{i}V_{i} = nRT_{i} to replace n and I got (after much simplifying)

V_{f} = p_{power}*$$\Delta$$t*R/[ p_{i}C_{p} ] + 1

I then plugged in all the numbers (using C_{p} of water = 4100 and got:

25*15*8.314/(1.11*10^5*4100) + 1 = 1.00000685 m^3

Which gives $$\approx$$ 1.000006.85 * 10^6 cm^3

I have a very hard time believing that is the answer -- it's just too large.

Any help pointing out the stupid thing(s) I did would be very appreciated!

THANKS!!!!!!

Last edited:

alphysicist
Homework Helper
Hi houseguest,

Did you type in the entire question? You used C_p of water in part of your solution, but I don't see where water appears in the problem.

Sorry, I forgot to mention that the container is filled with water.

alphysicist
Homework Helper
I hope I'm not misunderstanding the problem. However one point is that if the container volume is increasing because it's filled with water and the water volume is increasing, then you cannot use PV=nRT. (That applies to ideal gases.) Instead use the formula for volume expansion of a solid or liquid.

Wow! I'm sorry, that was definitely one of my stupid things. Thanks for pointing that out! The container is filled with a monatomic gas. So I should be using C_p for that.
So, since C_p = R + C_v and C_v = 3/2 * R ( for a monatomic gas) then C_p = 5/2 * R

So that would give ( with the R's canceling)

25*15/(1.11*10^5 * 5/2) + 1 = 1.00135 m^3 = 1.00135 * 10^6 cm^3

However, this (I would think) is still too large. Any ideas?