If there is a 10% decrease in the atmospheric pressure at a hill compa...
DP/dT=L/T(V2-V1)
dT=dp*T*(V2-V1)/L
=0.9*1.01*10^⁵*373*1.2/2270*10^³
=18degree celsius
( Here used Claudius clapreyon equation)
10�crease means 90% hence getting 0.9*1.01*10^5
T=373K
L=2270kj/kg converted to 2270*10^3
V2-V1(given in question)=1.2
If there is a 10% decrease in the atmospheric pressure at a hill compa...
Change in Atmospheric Pressure and Boiling Point of Water
Introduction:
Atmospheric pressure plays a significant role in determining the boiling point of liquids. The boiling point of water, for instance, is influenced by changes in atmospheric pressure. In this case, we will explore the relationship between a 10% decrease in atmospheric pressure at a hill and the corresponding change in the boiling point of water.
Understanding the Concept:
The boiling point of a liquid is the temperature at which its vapor pressure is equal to the atmospheric pressure. When the atmospheric pressure decreases, the vapor pressure required for boiling is also reduced, resulting in a lower boiling point.
Determining the Change in Boiling Point:
To calculate the change in boiling point, we need to consider the latent heat of vaporization and the change in specific volume at the boiling point.
1. Latent Heat of Vaporization:
The latent heat of vaporization (L) for water is given as 2270 kJ/kg. This represents the amount of energy required to convert one kilogram of water into vapor at its boiling point.
2. Change in Specific Volume:
The change in specific volume (Δv) at the boiling point of water is given as 1.2 m²/kg. This indicates the change in volume per unit mass when water undergoes a phase change from liquid to vapor at its boiling point.
Calculating the Change in Boiling Point:
The change in boiling point (∆Tb) can be determined using the equation:
∆Tb = (L * ∆v) / (R * Tb)
Where:
∆Tb = Change in Boiling Point
L = Latent Heat of Vaporization
∆v = Change in Specific Volume
R = Universal Gas Constant (8.314 J/(mol·K))
Tb = Boiling Point Temperature (in Kelvin)
Substituting the Given Values:
In this case, we are given a 10% decrease in atmospheric pressure. However, the specific atmospheric pressure values are not provided. We can still proceed with the calculations assuming a reference atmospheric pressure at sea level (P0).
Let's assume the boiling point of water at sea level is 100°C (373 K). We can now calculate the change in boiling point (∆Tb) using the provided data.
∆Tb = (L * ∆v) / (R * Tb)
∆Tb = (2270 kJ/kg * 1.2 m²/kg) / (8.314 J/(mol·K) * 373 K)
Calculating the Change in Boiling Point:
By performing the necessary calculations, we can find the value of ∆Tb.
∆Tb = (2270 kJ/kg * 1.2 m²/kg) / (8.314 J/(mol·K) * 373 K)
∆Tb ≈ 8.21 K
Hence, there will be an approximate 8.21°C decrease in the boiling point of water due to a 10% decrease in atmospheric pressure at a hill compared to sea level.
Conclusion:
A decrease in atmospheric pressure at a hill leads to a decrease in the boiling point of water. The change in boiling point