Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4 when heated by 1^ C. Its initial temperature

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

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Given: The pressure of an ideal gas in a closed vessel increases by $0.4\%$ when the temperature is increased by $1^\circ \text{C}$ .

Find: The initial temperature.

For an ideal gas,

PV = nRT

Since the vessel is closed, $V$ and $n$ remain constant. Therefore,

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

The pressure increases by $0.4\%$ , so

P_2 = P_1 \times 1.004

The temperature is increased by $1^\circ \text{C}$ , which is the same as an increase of $1 \, \text{K}$ , so

T_2 = T_1 + 1

Substitute these into the pressure-temperature relation:

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

Cancelling $P_1$ from both sides,

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

Cross-multiplying,

T_1 + 1 = 1.004T_1

So,

1 = 0.004T_1

Hence,

T_1 = \frac{1}{0.004} = 250 \, \text{K}

Therefore, the initial temperature is $250 \, \text{K}$ . The correct option is C.

Given: Pressure increases by 0.4 for a temperature rise of 1 K at constant volume. Find: The initial temperature. At constant volume for an ideal gas, pressure is directly proportional to absolute temperature: P T So the fractional change satisfies PP = TT Using 0.4100 = 1T we get 0.004 = 1T Therefore, T = 250 K This shortcut works because for a closed rigid vessel, V and n are constant, making PT constant. Hence the correct option is C.

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