Question

For a thermodynamic system, the coefficient of volume expansion \(\beta=\frac{1}{V}(\frac{∂V}{∂T})_P\) and compressibility \(K=-\frac{1}{V}(\frac{∂V}{∂P})_T\) , where V, T, and P are respectively the volume, temperature, and pressure. Considering that \(\frac{dV}{V}\) is a perfect differential, we get

• A. \((\frac{∂\beta}{∂P})_T=(\frac{∂K}{∂T})_P\)
• B. \((\frac{∂\beta}{∂T})_P=-(\frac{∂K}{∂P})_T\)
• C. \((\frac{∂\beta}{∂P})_T=-(\frac{∂K}{∂T})_P\)
• D. \((\frac{∂\beta}{∂T})_P=(\frac{∂K}{∂P})_T\)

Answer 1

The Correct Option is C

The question provides us with expressions for the coefficient of volume expansion and compressibility for a thermodynamic system. We are asked to find a relationship based on these expressions.

Given:

• The coefficient of volume expansion: \(\beta = \frac{1}{V}\left(\frac{∂V}{∂T}\right)_P\)
• The compressibility: \(K = -\frac{1}{V}\left(\frac{∂V}{∂P}\right)_T\)

To find the relationship, we start by considering that \(\frac{dV}{V}\) is a perfect differential. The total differential change in volume can be expressed using the following differential form:

\(dV = \left(\frac{∂V}{∂P}\right)_T dP + \left(\frac{∂V}{∂T}\right)_P dT\)

Dividing this equation by volume \(V\) gives:

\(\frac{dV}{V} = \frac{1}{V}\left(\frac{∂V}{∂P}\right)_T dP + \frac{1}{V}\left(\frac{∂V}{∂T}\right)_P dT\)

Substitute the expressions for \(\beta\) and \(K\):

\(\frac{dV}{V} = -K dP + \beta dT\)

Now, apply the condition that \(\frac{dV}{V}\) is a perfect differential, which implies that the mixed second derivatives satisfy the following equality:

\(\left(\frac{∂}{∂T} \left(-K\right)\right)_P = \left(\frac{∂}{∂P} \beta \right)_T\)

This leads to:

\(\left(\frac{∂K}{∂T}\right)_P = -\left(\frac{∂\beta}{∂P}\right)_T\)

Therefore, the correct answer is:

\((\frac{∂\beta}{∂P})_T=-(\frac{∂K}{∂T})_P\)

This shows that the derivative of the coefficient of volume expansion with respect to pressure at constant temperature is the negative of the derivative of compressibility with respect to temperature at constant pressure.