Question

A gas of certain mass filled in a closed cylinder at a pressure of $3.23\,\text{kPa}$ has temperature $50^\circ$C. The gas is now heated to double its temperature. The modified pressure is ___ Pa.

Hint:

For gases at constant volume, pressure is directly proportional to absolute temperature.

Answer 1

Correct Answer: 7

Step 1: Writing initial data.
\[ P_1 = 3.23\,\text{kPa} = 3230\,\text{Pa} \] \[ T_1 = 50^\circ\text{C} = 323\,\text{K} \] Step 2: Finding final temperature.
Since temperature is doubled:
\[ T_2 = 2T_1 = 646\,\text{K} \] Step 3: Using Gay-Lussac’s law (constant volume).
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] Step 4: Calculating final pressure.
\[ P_2 = \frac{P_1 T_2}{T_1} = \frac{3230 \times 646}{323} \] \[ P_2 = 6460\,\text{Pa} \approx 7 \times 10^{3}\,\text{Pa} \] Step 5: Final conclusion.
The modified pressure is approximately $7 \times 10^{3}\,\text{Pa}$.