`((dG)/(dp))_(T)=`

Options

A. `V`
B. `S`
C. `-S`
D. `-V`

Answer

To solve the question \(\left(\frac{dG}{dp}\right)_{T}\), we need to understand the relationship between Gibbs free energy (G), pressure (p), and temperature (T). Let's break it down step by step.

### Step 1: Understand Gibbs Free Energy
Gibbs free energy is defined by the equation:
\[ G = H - TS \]
where \(H\) is enthalpy, \(T\) is temperature, and \(S\) is entropy.

### Step 2: Differentiate Gibbs Free Energy
To find \(\left(\frac{dG}{dp}\right)_{T}\), we need to differentiate \(G\) with respect to pressure \(p\) while keeping temperature \(T\) constant. We can express the differential of \(G\) as:
\[ dG = dH - TdS - SdT \]
Since we are keeping \(T\) constant, \(dT = 0\), so the equation simplifies to:
\[ dG = dH - TdS \]

### Step 3: Use the Enthalpy Relation
Enthalpy \(H\) can be expressed in terms of internal energy \(U\) and pressure-volume work:
\[ H = U + pV \]
Differentiating \(H\) gives:
\[ dH = dU + pdV + Vdp \]

### Step 4: Substitute into the Differential
Now substituting \(dH\) into the equation for \(dG\):
\[ dG = (dU + pdV + Vdp) - TdS \]
This can be rearranged to:
\[ dG = dU + pdV + Vdp - TdS \]

### Step 5: Express Internal Energy Differential
Using the first law of thermodynamics, we know that:
\[ dU = TdS - pdV \]
Substituting this into the equation for \(dG\):
\[ dG = (TdS - pdV) + pdV + Vdp - TdS \]
This simplifies to:
\[ dG = Vdp \]

### Step 6: Take the Derivative
Now, taking the derivative of \(G\) with respect to \(p\) at constant \(T\):
\[ \left(\frac{dG}{dp}\right)_{T} = V \]

### Conclusion
Thus, the value of \(\left(\frac{dG}{dp}\right)_{T}\) is equal to the volume \(V\).

### Final Answer
\[ \left(\frac{dG}{dp}\right)_{T} = V \]