- #1

VincentweZu

- 13

- 0

## Homework Statement

the diagram above of pressure P versus volume V shows the expansion of 2.0 moles of a monatomic ideal gas from state A to state B. As shown in the diagram P

_{A}= P

_{B}= 600 N/m

^{2}, V

_{A}= 3.0 m

^{3}and V

_{B}= 9.0 m

^{3}.

a)

i) Calculate the work done by the gas as it expands

ii) Calculate the change in internal energy of the gas as it expands.

iii) Calculate the heat added to or removed from the gas during this expansion.

## Homework Equations

W = PΔV

U = [itex]\frac{3}{2}[/itex]nRT

PV = nRT

## The Attempt at a Solution

a)

i) W = PΔV

W = 600(9.0 - 3.0)

W = - 3600J

I put negative to indicate work done by system

ii) Here is where I have problems, I'm not sure how to calculate internal energy without given temperature. Here is my failed attempt.

U

_{A}= [itex]\frac{3}{2}[/itex]nRT

_{A}

Since PV = nRT

U

_{A}= [itex]\frac{3}{2}[/itex]P

_{A}V

_{A}

ΔU = U

_{B}- U

_{A}

= [itex]\frac{3}{2}[/itex]P

_{B}V

_{B}- [itex]\frac{3}{2}[/itex]P

_{A}V

_{A}

Since P

_{A}= P

_{B}= P

= [itex]\frac{3}{2}[/itex]P(V

_{B}- V

_{A})

= [itex]\frac{3}{2}[/itex]PΔV

= [itex]\frac{3}{2}[/itex]W

= [itex]\frac{3}{2}[/itex](-3600)

= -5400J

I am not sure that this is correct because I don't know if I am allowed to rearrange the equations like that, although it is correct algebraically, I'm not sure if it is correct physically.

I am unsure of whether the change in internal energy is supposed to be negative or positive, negative makes sense to me because work is being done by the gas and thus the internal energy should decrease.

iii) ΔU = Q + W

-5400 = Q -3600

Q = -1800J

Now I know there is something wrong since heat cannot be lost by the gas in an isobaric process.