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Experimentally one finds that for a rubber band (∂J/ ∂L) |T = (aT/ L0)*( 1 + 2 ( L0/ L)^3) and (∂J/ ∂T

)| L = (aL/ L0)( 1 − ( L0/ L)^3) ,

where T is temperature, J is the tension, a = 1.0 × 103dyn/K, and L0 = 0.5m is the length of the band when no tension is applied. The mass M of the rubber band is held fixed throughout. • Compute (∂L/∂T)J and discuss its physical meaning. • Show that dJ is an exact differential. • Calculate J and hence determine the equation of state.

Top Answer

*(dL/dT) J = - (L/T)[1-(L 0 /L) 3 ]/[1 + 2(L 0 /L) 3 ] *dJ = d[aT(L/L... View the full answer

IMG_20191015_055746.jpg

We have the chain rule for partial differentiation as-
7 (2 2)
= - (85/ OT ) 2/13 /L ) T
- ( a2 /Lo ) ( 1 - (LO/L ) 3 )
( GT / LO ) ( 1 + 2 ( Lo/ 2 ) s )
[1 - ( Lo/ 2 ) 3 )
[1 + 2 ( Lo/ 2 ) 3
OTJ -...

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