Question

how much work must be done on a systen to decrease its volume from 10.0 L to 2.0 L by exerting a constant pressure of 6.0 atm?

Answer 1

So technically, ΔV ..... and ΔBANANAS....ΔBubbles....ΔChocolate.... ΔANYthing, will always be Final-Initial.

.....If I am decreasing its INITIAL Vol (10.0 L) to its FINAL Vol (2.0 L)

ΔV∴ would be (2.0 L - 10.0 L) → ΔV= -8.0 L

in Addition, the SYSTEM is EXERTING a Constant Pressure... this means it is LOOSING 6.0 atm of Constant Pressure....

∴ P: -6.0 atm

**all together now**

→Work= Pressure * ΔVolume ⇒ (-6.0 atm)(-8.0 L) ⇒ 48 L*atm Work done ON the System where (+) above symbolizes this.

• Finally: 1 L* atm = 101.325 Joule. (Double check that they are not asking for the answer in kJ... which would then need to be converted further.) Otherwise,
• (48 L*atm)(101.325 J) = 4863.6 J  
• ***ALWAYS CHECK YOUR GIVEN SIG FIGS!!*** (round to 2 in this case)
• ∴ the Final Answer should read:  4.9 *10³ J work
• (thats a 3 superscript btw :-)

.....yes, it may seem nit-picky to negate these things, if the answer is the same... but trust me, as you progress through, you will definitely not want to forget what signs means what, and when they need to be changed from the information written to depict what is happening in/on/within/from the SYSTEM (<-- whose view we will always take.... because its got all the Bananas) :-P

Answer 2

Work done = P * ΔV = 6 * (10-2) = 48 atm l