Question

A block of metal whose volume
expansivity is 5.0 × 10−5 °C−1 and
isothermal compressibility is 1.2 ×
10−6 atm−1 has volume 5 litre at 1
atm and 20 °C. On applying
pressure, its temperature rises by
12°C and volume increases by 0.5
cm3. Calculate the pressure applied.

Answer 1

Here the expansion and compression both take place, making it interesting.

 

 Due to increase in pressure the volume decreases.  Due to increased temperature volume increases.  The effect of increasing temperature is more than that of the pressure.  So the net increase is 0.5 cc as given, in the block of the metal.

Expected expansion in volume due to ΔT = 12° C is 
  = 5000 cc * 5.0 * 10⁻⁵ / ⁰C * 12°C

  = 3.0 cc

But actual increase = 0.50 cc

So compression ΔV in volume V due to ΔP = 3 – 0.50 = 2.50 cc

Compressibility is defined as β_T = - 1/V * ΔV / ΔP

         Given  β_T  = 1.2 * 10⁻⁶ / atm

 ΔP = 2.5 / [ 5000 * 1.2 * 10⁻⁶) = 416.67 atm

Pressure applied = P + ΔP = 417.67 atm       ans.

Formula:

$Compressibility= \beta_T=-\frac{1}{V}\frac{\Delta V}{\Delta P}$