Discussion
The normal boiling point of a liquid is 200 K. If the liquid behaves as ideal liquid at 76 mm, then its boiling point will be close to : (Given : \(\dfrac{\Delta H}{R} \approx 2303\,\text{K}\))
Answer: A
The boiling point at 76 mm is close to 167 K.
Use the Clausius-Clapeyron equation between the normal boiling point (760 mm, 200 K) and the new pressure (76 mm, T2):
\[ \begin{aligned} \ln\frac{p_{2}}{p_{1}} &= -\frac{\Delta H}{R} \\ &\quad \times \left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right) \end{aligned} \]
\[ \begin{aligned} \ln 0.1 &= -2.303 \\ -2.303 &= -2303\left(\frac{1}{T_{2}}-\frac{1}{200}\right) \\ \frac{1}{T_{2}} &= 0.005 + 0.001 = 0.006 \\ T_{2} &= 166.7\ \text{K} \end{aligned} \]
Lower pressure must lower the boiling point, so 200 K (unchanged) and 440 K (higher) are impossible, and 20 K is what you get by scaling the temperature in proportion to the pressure, which the equation does not allow.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development