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Step-by-step Solution
Step 1: Express the given proportionality
We are given that the pressure of a gas varies as the cube of its absolute temperature during an adiabatic process. Mathematically:
$$P \propto T^3 \quad \Longrightarrow \quad P\,T^{-3} = \text{constant} \,.\,$$
Step 2: Write the standard adiabatic relation between pressure and temperature
For an adiabatic process of an ideal gas, there is a known relation:
$$P^{\,1 - \gamma}\, T^{\,\gamma} = \text{constant} \,,$$
where $$\gamma = \frac{C_p}{C_V}.$$
Rewriting this, we can isolate $P$ to get:
$$P \, T^{\frac{\gamma}{1 - \gamma}} = \text{constant} \,.$$
Step 3: Equate exponents from the given and standard relations
From Step 1, we have $P\,T^{-3} = \text{constant}$. From Step 2, we have $P\,T^{\frac{\gamma}{\,1 - \gamma}} = \text{constant}$. Equating the exponents on $T$:
$$\frac{\gamma}{1 - \gamma} = -\,3 \,.$$
Step 4: Solve for $$\gamma$$
We solve the equation:
$$\frac{\gamma}{1 - \gamma} = -3 \,.$$
Multiply both sides by $1 - \gamma$:
$$\gamma = -\,3 \,\bigl(1 - \gamma\bigr) \,.$$
Expand the right-hand side:
$$\gamma = -\,3 + 3\,\gamma \,.$$
Rearrange terms:
$$\gamma - 3\,\gamma = -\,3 \quad \Longrightarrow \quad -\,2\,\gamma = -\,3 \quad \Longrightarrow \quad \gamma = \frac{3}{2} \,.$$
Step 5: Conclude the ratio $C_p/C_V$
Since $$\gamma = \frac{C_p}{C_V} \,,$$ we obtain
$$\frac{C_p}{C_V} = \frac{3}{2} \,.$$
Final Answer
The ratio $$\frac{C_p}{C_V}$$ for the gas is $$\frac{3}{2}\,.$$
