Rankers Physics

Thermodynamics: Practice Problem & Solution

A given volume of air at 17°C exerts a pressure of 870 mm of mercury. The temperature at which the pressure becomes 1800 mm of Hg is :
300 K
400 K
500 K
600 K

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Thermodynamics. Understanding the underlying formula is key to arriving at the correct answer below:

To solve this, we can use Gay-Lussac's Law, which states:

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

Given:
- Initial pressure \( P_1 = 870 \, \text{mm Hg} \),
- Final pressure \( P_2 = 1800 \, \text{mm Hg} \),
- Initial temperature \( T_1 = 17^\circ \text{C} = 17 + 273 = 290 \, \text{K} \).

Rearrange to find \( T_2 \):

\[
T_2 = \frac{P_2 \times T_1}{P_1}
\]

Substitute the values:

\[
T_2 = \frac{1800 \times 290}{870}
\]

Calculating this:

\[
T_2 = 600 \, \text{K}
\]

So, the temperature at which the pressure becomes 1800 mm of Hg is  600 K.

Leave a Reply

Your email address will not be published. Required fields are marked *