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

Coefficient of volume expansion = ΔV / (V * ΔT) = 5.0 * 10⁻⁵ /°C
Isothermal compressibility = -1/V * ΔV/ΔP = - 1.2 * 10⁻⁶ / atm = - 0.12 / Pa
V1 = 5 litre = 0.005 m³,     P1 = 1 atm = 10⁵ Pa,    T1 = 20°C
V2 = 5.0005 litre,  ΔV = 5 * 10⁻⁷ m³
T2 = 32°C ,  ΔT = 12°C,
      P2 = ?    ΔP = ?

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) = 1/v * (dV/dp)

         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.