Question

A gas at a pressure of 2 atm is heated from 25 °C to 323 °C and simultaneously compressed to \(\frac{2^{rd}}{3}\) of its original value. Then the final pressure is

• A. 1.33 atm
• B. 6 atm
• C. 2 atm
• D. 4 atm

Answer 1

The Correct Option is B

Given:

• Initial pressure \( P_1 = 2 \) atm
• Initial temperature \( T_1 = 25^\circ C = 298 \) K
• Final temperature \( T_2 = 323^\circ C = 596 \) K
• Volume remains constant

 

Objective: Find the final pressure \( P_2 \).

Using the combined gas law: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] Rearranging to solve for \( P_2 \): \[ P_2 = P_1 \times \frac{T_2}{T_1} \]

Substituting the given values: \[ P_2 = 2 \times \frac{596}{298} \approx 3.99 \text{ atm} \]

Note: The calculated pressure \( P_2 \approx 4 \) atm does not match any of the provided options. The stated correct answer is D. 6 atm, which suggests there might be additional information or a different process involved.

Answer 2

We can use the combined gas law, which is given by:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:

• P₁ = initial pressure = 2 atm
• T₁ = initial temperature = 25°C = 298 K
• P₂ = final pressure (unknown)
• V₁ = initial volume (considered 1 for simplicity)
• V₂ = final volume = (2/3)V₁
• T₂ = final temperature = 323°C = 596 K

Rearranging the combined gas law to solve for P₂:

\[ P_2 = P_1 \times \left( \frac{V_1}{V_2} \right) \times \left( \frac{T_2}{T_1} \right) \]

Substitute the known values:

\[ P_2 = 2 \, \text{atm} \times \left( \frac{1}{\frac{2}{3}} \right) \times \left( \frac{596}{298} \right) = 6 \, \text{atm} \]

Thus, the final pressure is 6 atm.