# Find work done on an isobaric system

• Feodalherren
In summary, the conversation discussed finding the work done on a 1kg box of Aluminum as it is warmed from 22C to 40C at a constant pressure of 1atm. The correct equation for work was determined to be W= -PΔV, where the change in volume is given by the linear coefficient of expansion of Aluminum. The final answer was found to be around 48mJ.
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

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.

Work is PΔV. What is the change of volume?

ehild

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

1 person
Oh yeah I totally missed that. Thanks.

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.

## What is an isobaric system?

An isobaric system is a thermodynamic system in which the pressure remains constant while the volume and temperature can vary.

## What is the formula for calculating work done on an isobaric system?

The formula for calculating work done on an isobaric system is W = P∆V, where W is the work done, P is the constant pressure, and ∆V is the change in volume.

## How is work done on an isobaric system represented on a graph?

On a pressure-volume (PV) diagram, work done on an isobaric system is represented by the area under the curve, which is equal to the product of the constant pressure and the change in volume.

## What are some real-life examples of isobaric systems?

Some real-life examples of isobaric systems include a gas cylinder with a constant pressure, a balloon being inflated at a constant pressure, and a steam engine with a constant pressure in the boiler.

## How does the work done on an isobaric system affect its internal energy?

The work done on an isobaric system increases its internal energy, as energy is transferred to the system in the form of work. This leads to an increase in temperature, as the internal energy is directly proportional to the temperature of the system.

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