Posted by **Sammy** on Saturday, July 13, 2013 at 2:18pm.

7 Moles of a gas initially at temperature 300 K is compressed adiabatically from a volume of 1000 cm3 to a volume of 316 cm3. To the nearest tenth of a kJ what is the work done by the piston? (It is a monatomic ideal gas.)

I have W = (3/2)nRT1(V1^(gamma-1) V2^(1-gamma) times -1)

I tried to set it up and I got 56.4 but the answer was 30.3. Can someone tell me what I did wrong and tell me how to properly calculate it? Thanks!

- College Physics -
**Elena**, Saturday, July 13, 2013 at 5:21pm
v= 7 moles,

T₁=300 K

V₁=1000cm³= 10⁻³ m³

V₂=0.316•10⁻³ m³

W= {(vRT₁)/(γ-1)} •[1-(V₁/V₂)^(γ-1)]=

γ=(i+2)/I =(3+2)/3 = 5/3

γ-1=5/3 -1 =2/3= 0.667

W=- {(vRT₁)/(γ-1)} •[1-(V₁/V₂)^(γ-1)]=

= - { 7•8.31•300/0.667}•[1- (10⁻³/0.316•10⁻³)^0.667] =

= 30252 J =30.252 kJ ≈ 30.3 kJ

- College Physics -
**Josh**, Thursday, July 25, 2013 at 9:50am
An ice skater spinning with outstretched arms has an angular speed of 5.0\({\rm rad/s}\) . She tucks in her arms, decreasing her moment of inertia by 19\({\rm \%}\) .

By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)

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