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 \( \left( \frac{\partial p}{\partial T} \right)_S \) 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 \( \beta = 1.6 \times 10^{-4} \, \text{K}^{-1} \). If the value of \( \left( \frac{\partial p}{\partial T} \right)_S \) at 270 K is \( N \times 10^7 \, \text{Pa/K} \), then the value of \( N \) is:

Hint:

Maxwell's relations are powerful tools for deriving thermodynamic quantities from known variables. Here, we use them to express the pressure-temperature derivative.

Answer 1

Correct Answer: 4.6

Step 1: Use Maxwell's relations.
From thermodynamics, Maxwell's relations give the following expression for \( \left( \frac{\partial p}{\partial T} \right)_S \): \[ \left( \frac{\partial p}{\partial T} \right)_S = \frac{\rho \beta}{c_p} \] Step 2: Substituting known values.
Using the values \( c_p = 2010 \, \text{J/kg-K} \), \( \rho = 10^3 \, \text{kg/m}^3 \), and \( \beta = 1.6 \times 10^{-4} \, \text{K}^{-1} \), we can calculate: \[ \left( \frac{\partial p}{\partial T} \right)_S = \frac{10^3 \times 1.6 \times 10^{-4}}{2010} = 4.60 \times 10^7 \, \text{Pa/K} \]