Question

When a monoatomic gas at a pressure $P$ is suddenly compressed to $\frac{1}{8}$ of its original volume, the pressure of the gas becomes}

• A. 8P
• B. 5P
• C. 3P
• D. 32P
• E. 16P

Hint:

The word "suddenly" is a standard keyword in physics problems to indicate an adiabatic process ($PV^\gamma = \text{const}$), while "slowly" indicates an isothermal process ($PV = \text{const}$).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A "sudden" compression implies an adiabatic process because there is no time for heat exchange with the surroundings.
Step 2: Key Formula or Approach:
For an adiabatic process, the relationship between pressure and volume is \(PV^\gamma = \text{constant}\).
Therefore, \(P_1 V_1^\gamma = P_2 V_2^\gamma\).
For a monoatomic gas, \(\gamma = \frac{5}{3}\).
Step 3: Detailed Explanation:
Given values:
Initial state: \(P_1 = P\), \(V_1 = V\).
Final state: \(V_2 = \frac{V}{8}\).
Set up the adiabatic equation:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \] \[ P (V)^{5/3} = P_2 \left(\frac{V}{8}\right)^{5/3} \] Rearrange to solve for \(P_2\):
\[ P_2 = P \left(\frac{V}{V/8}\right)^{5/3} \] \[ P_2 = P (8)^{5/3} \] Rewrite 8 as a power of 2: \(8 = 2^3\).
\[ P_2 = P ((2^3)^{1/3})^5 \] \[ P_2 = P (2)^5 \] \[ P_2 = 32P \] Step 4: Final Answer:
The final pressure is 32P.