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Question

Pressure of an ideal gas, contained in a closed vessel, is increased by $0.4\%$ when heated by $1^\circ C$. Its initial temperature must be:

The correct answer is
$250 \text{ K}$

Ideal Gas Law at Constant Volume

For an ideal gas contained in a closed vessel, the volume ($V$) remains constant. The ideal gas law states $PV = nRT$. Since the amount of gas ($n$) and the universal gas constant ($R$) are constant, we have $P \propto T$ when $V$ is constant.

This relationship can be expressed as a ratio:

$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $

where $P_1, T_1$ are the initial pressure and absolute temperature, and $P_2, T_2$ are the final pressure and absolute temperature.

Calculating Initial Temperature

Let the initial absolute temperature be $T_1$ (in Kelvin).

The temperature increases by $1^\circ C$. A change of $1^\circ C$ is equivalent to a change of $1 \text{ K}$ in absolute temperature. So, the final absolute temperature is:

$ T_2 = T_1 + 1 \text{ K} $

The pressure increases by $0.4\%$. The final pressure $P_2$ is:

$ P_2 = P_1 + 0.004 P_1 = 1.004 P_1 $

Step-by-Step Derivation

  1. Substitute the expressions for $P_2$ and $T_2$ into the constant volume gas law ratio:

    $ \frac{P_1}{T_1} = \frac{1.004 P_1}{T_1 + 1} $

  2. Since $P_1$ is non-zero, we can cancel it from both sides:

    $ \frac{1}{T_1} = \frac{1.004}{T_1 + 1} $

  3. Cross-multiply:

    $ T_1 + 1 = 1.004 T_1 $

  4. Rearrange the terms to solve for $T_1$:

    $ 1 = 1.004 T_1 - T_1 $

    $ 1 = (1.004 - 1) T_1 $

    $ 1 = 0.004 T_1 $

  5. Calculate the initial temperature $T_1$:

    $ T_1 = \frac{1}{0.004} = \frac{1000}{4} $

    $ T_1 = 250 \text{ K} $

The initial temperature of the ideal gas must be $250 \text{ K}$.

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