in a closed system at 40 C, a liquid has a vapor pressure of 50 KPa. the liquid's normal boiling point could be
(1)10 C
(2)30 C
(3)40 C
(4)60 C
60
It's 60 C for the reason of its must be higher then forty
Doesn't one need the heat of vaporization, then use the Clausius-Clapeyron equation?
To determine the liquid's normal boiling point, we need to compare the given vapor pressure at 40°C with the standard atmospheric pressure, which is approximately 101.3 kPa. If the vapor pressure is lower than the standard atmospheric pressure, the liquid will boil at a temperature higher than its normal boiling point. On the other hand, if the vapor pressure is higher than the standard atmospheric pressure, the liquid will boil at a temperature lower than its normal boiling point.
In this case, the given liquid has a vapor pressure of 50 kPa at 40°C. As 50 kPa is lower than 101.3 kPa, we can conclude that the liquid will boil at a temperature higher than its normal boiling point. Therefore, option (1), 10°C, can be eliminated.
Now, we need to consider the remaining options: 30°C, 40°C, and 60°C.
At the normal boiling point, the vapor pressure of a liquid is equal to the standard atmospheric pressure. Since the given liquid's vapor pressure is below the standard atmospheric pressure at 40°C, it indicates that the boiling point is higher than 40°C. Therefore, we can eliminate option (3), 40°C.
Now, we are left with two options: 30°C and 60°C. To determine which one is correct, let's analyze the given information. At 40°C, the liquid's vapor pressure is 50 kPa. This means that at a higher temperature, such as 60°C, the vapor pressure of the liquid would be even greater.
Since the liquid's vapor pressure is below the standard atmospheric pressure at 40°C and would be higher at 60°C, we can conclude that the liquid will boil at a temperature lower than 60°C. Therefore, the correct answer is option (2), 30°C.
In summary, based on the given information, the liquid's normal boiling point could be 30°C.