Question

4. A rubber band of dimensions 10 mm x 2 mmx 200 mm is stretched 5 times its original length at 20�C. Calculate the heat gained or lost. The number of chains per unit m^3 is 1.9 x 10^26.Where w is the work per unit volume. Delta U = delta W + delta Q (U = internal energy of system, W = total work performed on the system. Q = heat added to the system). Hint: you need to assume that there is no change to the internal energy of the system. and that there is no change in temperature.

Answer 1

Dimensions of rubber band = 10 mm x 2 mm x 200 mm.

Volume of rubber band = 4000 mm³

= 4.0x10^-6 m³

Temperature, T = 20 oC = 293 K

Chains / unit m³ = 1.9x10²⁶

Work done per unit volume, w = -1/2 $\nu$kT ($\lambda$² + 2/$\lambda$ - 3)

= -1/2 * 1.9x10²⁶ * 1.38x10^-23 * 293 (5² + 2/5 - 3)

= -1/2 * 1.9x10²⁶ * 1.38x10^-23 * 293 * 22.4

= - 4.51x10⁶ J / m³

Total work done, W = - 4.51x10⁶ J / m³ * 4.0x10^-6 m³

= - 18.04 J

$\Delta$U = $\Delta$W + $\Delta$Q

Since, $\Delta$U = 0

$\Delta$Q = - $\Delta$W

$\Delta$Q = 18.04 J

So, Heat lost = $\Delta$Q = 18.04 J