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: Identify the process.
The gas is compressed adiabatically (no heat exchange), so it obeys Poisson's relation $PV^{\gamma} = \text{constant}$, with $\gamma = \tfrac{7}{5} = 1.4$.
Step 2: Switch from volume to density.
For a fixed mass, volume and density are inversely related, $V \propto 1/\rho$. Substituting into $PV^{\gamma} = $ constant gives $$\frac{P}{\rho^{\gamma}}\ \text{is not constant; instead}\quad P \propto \rho^{\gamma}.$$ Hence $\dfrac{P_2}{P_1} = \left(\dfrac{\rho_2}{\rho_1}\right)^{\gamma}.$
Step 3: List the known values.
$P_1 = p$, and the density doubles, so $\rho_2 = 2\rho_1$, giving the ratio $\rho_2/\rho_1 = 2$.
Step 4: Substitute into the density relation.
$$P_2 = p \times (2)^{1.4}.$$
Step 5: Evaluate $2^{1.4}$.
Write $2^{1.4} = 2 \times 2^{0.4}$. Since $2^{0.4} = \sqrt[5]{4} \approx 1.32$, we get $2^{1.4} \approx 2 \times 1.32 = 2.64.$
Step 6: Write the final pressure.
Therefore $P_2 \approx 2.63\,p$, slightly above double the original pressure, as expected when compression also raises the temperature.
\[ \boxed{P_2 \approx 2.63\,p} \]