Question

Starting with the equation \( TdS = dU + p dV \) and using the appropriate Maxwell's relation along with the expression for heat capacity \( C_p \) (see useful information), the derivative \( \left( \frac{\partial p}{\partial T} \right)_V \) for a substance can be expressed in terms of its specific heat \( c_p \), density \( \rho \), coefficient of volume expansion \( \beta \), and temperature \( T \). For ice, \( c_p = 2010 \, \text{J/kg.K} \), \( \rho = 10^3 \, \text{kg/m}^3 \), and \( \beta = 1.6 \times 10^{-4} \, \text{1/K} \). If the value of \( \left( \frac{\partial p}{\partial T} \right)_V \) at 270 K is \( N \times 10^7 \, \text{Pa/K} \), then the value of \( N \) is:

Hint:

Maxwell's relations allow you to express various thermodynamic derivatives in terms of specific heat, density, and other material properties.

Answer 1

Step 1: Maxwell's relation.
We start with the equation \( TdS = dU + p dV \). Using Maxwell's relation, we know that: \[ \left( \frac{\partial p}{\partial T} \right)_V = \beta c_p \rho \] Substituting the values for \( c_p \), \( \rho \), and \( \beta \), we get: \[ \left( \frac{\partial p}{\partial T} \right)_V = (1.6 \times 10^{-4}) \times (2010) \times (10^3) \] Step 2: Perform the calculation.
Calculating this gives: \[ \left( \frac{\partial p}{\partial T} \right)_V = 3.216 \times 10^7 \, \text{Pa/K} \] Step 3: Conclusion.
Thus, the value of \( N \) is 3.2.