Question

An ideal gas at pressure 'p' is adiabatically compressed so that its density becomes twice that of the initial. If $\gamma = \frac{c_p}{c_v} = \frac{7}{5}$, then final pressure of the gas is

• A. p
• B. 2p
• C. $\frac{7}{5}$ p
• D. 2.63p

Hint:

In any compression process, the final pressure must increase, which allows you to quickly eliminate options (A) and (C). Since adiabatic processes are steeper than isothermal processes on a P-V diagram due to the $\gamma$ exponent, the pressure must increase by a factor greater than the density ratio ($2^1 = 2$). This leaves $2.63p$ as the only mathematically sound choice.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem describes an ideal gas undergoing an adiabatic compression process that doubles its mass density. Given the specific heat ratio ($\gamma$), we need to find the final equilibrium pressure in terms of the initial pressure $p$.

Step 2: Key Formula or Approach:
For a reversible adiabatic thermodynamic process, the relationship between pressure ($P$) and volume ($V$) follows Poisson's law:
$$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$$ Density ($\rho$) is inversely proportional to volume ($V = \frac{M}{\rho}$) for a fixed mass of gas. Substituting this into Poisson's law allows us to express the relationship directly in terms of density:
$$P_1 \left(\frac{1}{\rho_1}\right)^{\gamma} = P_2 \left(\frac{1}{\rho_2}\right)^{\gamma} \implies \frac{P_2}{P_1} = \left(\frac{\rho_2}{\rho_1}\right)^{\gamma}$$

Step 3: Detailed Explanation:
Let's organize the parameters given in the problem statement:
Initial state pressure: $P_1 = p$, and initial density is $\rho_1$.
Final state density: $\rho_2 = 2\rho_1$.
Adiabatic index exponent: $\gamma = \frac{7}{5} = 1.4$.
Substitute these parameters into the density-based adiabatic relationship:
$$\frac{P_2}{p} = \left(\frac{2\rho_1}{\rho_1}\right)^{1.4}$$ The density variables cancel out inside the parentheses:
$$P_2 = p \times (2)^{1.4}$$ Now, we calculate the numerical value of $2^{1.4}$. We can rewrite this expression fractionally as $2^{7/5}$:
$$2^{1.4} = 2^{1} \times 2^{0.4} = 2 \times 2^{2/5} = 2 \times \sqrt[5]{4} \approx 2 \times 1.3195 = 2.639$$ Rounding to two decimal places matches the value provided in the options:
$$P_2 \approx 2.63p$$ This calculation determines the final pressure after compression.

Step 4: Final Answer:
The final pressure of the gas is 2.63p, which corresponds to option (D).