During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of \(\frac{C_p}{C_v}\) for the gas is :
- \(\frac{5}{3}\)
- \(\frac{9}{7}\)
- \(\frac{3}{2}\)
- \(\frac{7}{5}\)
The Correct Option is C
Approach Solution - 1
Given that:
\[ P \propto T^3, \]
where \( P \) is the pressure and \( T \) is the absolute temperature.
Step 1: Using the Ideal Gas Law
From the ideal gas law, we have:
\[ \frac{PV}{T} = nR = \text{constant}. \]
Therefore:
\[ P \propto \frac{T}{V}. \]
Step 2: Relating Pressure and Temperature
Given that:
\[ P \propto T^3, \]
we can write:
\[ P = kT^3, \]
where \( k \) is a proportionality constant.
Step 3: Applying the Adiabatic Process Equation
For an adiabatic process, the relation is given by:
\[ PV^\gamma = \text{constant}, \]
where \( \gamma = \frac{C_P}{C_V} \) is the adiabatic index.
Step 4: Comparing the Relations
From the given proportionality:
\[ P \propto T^3 \quad \text{and} \quad P \propto V^{-\gamma}. \]
Equating the exponents:
\[ \gamma = 3. \]
Thus, the ratio of \( \frac{C_P}{C_V} \) is:
\[ \frac{C_P}{C_V} = \gamma = \frac{7}{5}. \]
Therefore, the correct answer is \( \frac{7}{5} \).
Approach Solution -2
For an adiabatic process, the relation between pressure (P) and temperature (T) is given as:
\[ P \propto T^3 \]
We need to find the ratio \( \frac{C_p}{C_v} \).
Step 2: Recall the adiabatic relation.
For an adiabatic process:
\[ P V^{\gamma} = \text{constant} \] and also, the ideal gas law gives: \[ P V = R T \]
where \( \gamma = \frac{C_p}{C_v} \).
Step 3: Eliminate volume (V).
From the gas law: \[ V = \frac{R T}{P} \]
Substitute in \( P V^{\gamma} = \text{constant} \):
\[ P \left(\frac{R T}{P}\right)^{\gamma} = \text{constant} \] \[ P^{1 - \gamma} T^{\gamma} = \text{constant} \]
Step 4: Express P in terms of T.
\[ P \propto T^{\frac{\gamma}{\gamma - 1}} \] Given that \( P \propto T^3 \), we can equate exponents:
\[ \frac{\gamma}{\gamma - 1} = 3 \]
Step 5: Solve for γ.
\[ \gamma = 3(\gamma - 1) \] \[ \gamma = 3\gamma - 3 \] \[ 2\gamma = 3 \] \[ \gamma = \frac{3}{2} \]
Step 6: Final Answer.
The ratio \( \frac{C_p}{C_v} \) for the gas is:
\[ \boxed{\frac{3}{2}} \]
Final Answer: \( \frac{3}{2} \)
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