Find work done on an isobaric system

[Feodalherren]

Homework Statement

A 1kg box of Aluminum initially at 22C is warmed at P=1atm so that its T increases to 40C. Find
a) The work done on the system

Homework Equations

W= -∫PdV

The Attempt at a Solution

The pressure is constant at 101,300Pa

D=m/V

So V=m/D

D of Al at room temp = 2700kg/m^3

Therefore

$W= -101,300Pa \frac{1kg}{2700kg/m^{3}}(1+24x10^{-6}(18))^{3}$

Where ΔT=18 and 24x10^-6 is the linear coefficient of expansion of Aluminum.

= -37.6J

Not the right answer, it should be around 48mJ. I can't see what I'm doing wrong here.

 

[ehild]

Work is PΔV. What is the change of volume?

ehild

 

[Chestermiller]

It should be

$W= -101,300Pa \frac{1kg}{2700kg/m^{3}}((1+24\times 10^{-6}(18))^{3}-1)$

because you forgot to subtract the initial volume. Essentially the same result, which retains only the linear term, is:

$W= -101,300Pa \frac{1kg}{2700kg/m^{3}}(3)(24\times 10^{-6}(18))$

Chet

 

[Feodalherren]

Oh yeah I totally missed that. Thanks.

 

[HalfLostAndSearching]

Your approach to finding the work done on the system is correct, but your calculation for the change in volume is incorrect. The linear coefficient of expansion for aluminum is given in units of 1/°C, not 1/m. So, the correct calculation should be:

ΔV = (2700kg/m^3)(1+24x10^-6/°C)(40°C-22°C)

= 2700kg/m^3(1+24x10^-6/°C)(18°C)

= 0.389m^3

Plugging this into the work equation, we get:

W = -101,300Pa(0.389m^3)

= -39.3J

This is closer to the expected answer of 48mJ, but it is still slightly off. This could be due to rounding errors or slight variations in the given values. However, the overall approach and method is correct.