Question

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

• A. \( 25^\circ \text{C} \)
• B. \( 2500 \, \text{K} \)
• C. \( 250 \, \text{K} \)
• D. \( 250^\circ \text{C} \)

Hint:

For an ideal gas in a closed vessel (isochoric process), the ratio of pressure to temperature (in Kelvin) is constant. Use the given percentage increase in pressure and the change in temperature to set up a proportion and solve for the initial temperature in Kelvin. Remember to always use Kelvin for gas law calculations involving temperature.

Answer 1

The Correct Option is C

The gas is contained in a closed vessel, so the volume remains constant. This is an isochoric process. For an ideal gas at constant volume, the pressure is directly proportional to the temperature (in Kelvin): \[ P \propto T \implies \frac{P}{T} = \text{constant} \implies \frac{\Delta P}{\Delta T} = \frac{P}{T} \] Let the initial pressure be \( P \) and the initial temperature be \( T \) (in Kelvin). 
The pressure increases by 0.4%, so the change in pressure \( \Delta P = 0.4% \text{ of } P = \frac{0.4}{100} P = 0.004 P \). The temperature is increased by \( 1^\circ \text{C} \), which is equal to \( 1 \, \text{K} \) change in Kelvin scale, so \( \Delta T = 1 \, \text{K} \). 
Substituting these values into the equation: \[ \frac{0.004 P}{1} = \frac{P}{T} \] \[ 0.004 = \frac{1}{T} \] \[ T = \frac{1}{0.004} = \frac{1000}{4} = 250 \, \text{K} \] The initial temperature of the gas is \( 250 \, \text{K} \). 
To convert this to Celsius, we use \( T(^\circ \text{C}) = T(K) - 273.15 \): \[ T(^\circ \text{C}) = 250 - 273.15 = -23.15^\circ \text{C} \] However, the options are given in Celsius and Kelvin, and \( 250 \, \text{K} \) is one of the options.