Question

If the pressure of a gas increases by 2% at constant volume, then its temperature

• A. increases by 2%
• B. decreases by 2%
• C. does not change
• D. decreases by 1%

Hint:

Gay-Lussac’s Law: At constant volume: \[ \fracPT = \textconstant \Rightarrow \frac\Delta PP = \frac\Delta TT \] If pressure increases by 2\%, temperature must also increase by 2\%.

Answer 1

The Correct Option is A

Using the ideal gas law: $PV = nRT$. At constant volume: \[ \frac{P}{T} = \text{constant} \Rightarrow \frac{P_1}{T_1} = \frac{P_2}{T_2} \] Given $P_2 = 1.02 P_1$, then: \[ \frac{P_1}{T_1} = \frac{1.02 P_1}{T_2} \Rightarrow T_2 = 1.02 T_1 \] So, temperature also increases by 2%. Thus, option (1) is correct.

Answer 2

Step 1: Understand the relationship between pressure and temperature at constant volume
According to Gay-Lussac's law (or Amontons' law), for a given amount of gas at constant volume, the pressure of the gas is directly proportional to its absolute temperature:
\[ P \propto T \quad \text{or} \quad \frac{P}{T} = \text{constant} \]

Step 2: Express the relationship mathematically
At constant volume, if the initial pressure and temperature are \( P_1 \) and \( T_1 \), and the final pressure and temperature are \( P_2 \) and \( T_2 \), then:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \implies \frac{P_2}{P_1} = \frac{T_2}{T_1} \]

Step 3: Calculate percentage change in temperature corresponding to pressure increase
Given that pressure increases by 2%, so:
\[ \frac{P_2}{P_1} = 1 + \frac{2}{100} = 1.02 \]
Since \( \frac{P_2}{P_1} = \frac{T_2}{T_1} \), the temperature also increases by 2%.

Step 4: Conclusion
Therefore, if the pressure of a gas increases by 2% at constant volume, its temperature also increases by 2%.