Question:medium

When the pressure of the gas contained in a closed vessel is increased by 2.5%, the temperature of the gas increases by 4 K . The initial temperature of the gas is

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$P \propto T \implies \frac{\Delta P}{P} = \frac{\Delta T}{T}$ for small changes.
Updated On: May 7, 2026
  • 80 K
  • 150 K
  • 160 K
  • 320 K
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The gas is contained in a closed vessel, which means its volume remains constant ($V = \text{constant}$).
According to Gay-Lussac's law, for a fixed mass of an ideal gas at constant volume, the pressure is directly proportional to its absolute temperature ($P \propto T$).
Step 2: Key Formula or Approach:
From $P \propto T$, we can write: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] For small percentage changes, we can also use the fractional change relation: \[ \frac{\Delta P}{P} = \frac{\Delta T}{T} \] Step 3: Detailed Explanation:
Let the initial pressure be $P$ and the initial temperature be $T$.
The pressure is increased by $2.5%$.
The fractional change in pressure is: \[ \frac{\Delta P}{P} = \frac{2.5}{100} = 0.025 \] The temperature increases by $4 \text{ K}$, so the change in temperature is: \[ \Delta T = 4 \text{ K} \] Using the fractional change formula (which is exact here because it is a direct linear proportionality): \[ \frac{\Delta P}{P} = \frac{\Delta T}{T} \] Substitute the known values: \[ 0.025 = \frac{4}{T} \] Rearrange the equation to solve for $T$: \[ T = \frac{4}{0.025} \] Multiply numerator and denominator by 1000 to remove the decimal: \[ T = \frac{4000}{25} \] \[ T = 160 \text{ K} \] Alternatively, using the exact state equations: $P_2 = P + 0.025P = 1.025P$ $T_2 = T + 4$ $\frac{P}{T} = \frac{1.025P}{T + 4} \implies 1.025T = T + 4 \implies 0.025T = 4 \implies T = 160 \text{ K}$.
Step 4: Final Answer:
The initial temperature of the gas is 160 K.
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