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.)

Question

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!

Answer 1

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

Answer 2

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.)