Important Numerical Questions for NET GATE: Atmosphere

1. Water vapor exert 6mb pressure at 27 Celsius. what is density of water ?

• Water vapor pressure: 6 mb (millibars). This is the partial pressure exerted by the water vapor in the air.
• Temperature: 27 °C (Celsius). This is the temperature at which the water vapor exists.

Formula

We’ll use the Ideal Gas Law to relate pressure, volume, temperature, and the number of moles of a gas:

• PV = nRT

Where:

• P = Pressure (in Pascals)  
• V = Volume (in cubic meters)  
• n = Number of moles  
• R = Ideal Gas Constant (8.314 J/(mol·K))
• T = Temperature (in Kelvin)  

We also need to know:

• Molar mass of water (H₂O) = 18 g/mol  

We can rearrange the Ideal Gas Law to solve for density:

1. Density = mass/volume  
2. mass = n * molar mass
3. From the Ideal Gas Law, n/V = P/RT
4. Therefore, Density = (P/RT) * molar mass

• Pressure:
  • 1 mb = 100 Pa  
  • 6 mb = 600 Pa
• Temperature:
  • T(K) = T(°C) + 273.15
  • 27 °C = 300.15 K

Calculation

Density = (P/RT) * molar mass Density = (600 Pa) / ((8.314 J/(mol·K)) * (300.15 K)) * (18 g/mol) Density = 0.00432 kg/m³  

The density of the water vapor is approximately 4.32 x 10⁻³ kg/m³.

2. On a particular day, the air temperature is measured to be 30 °C, and the dew point is 20 °C. Calculate the actual vapor pressure. If the saturation vapor pressure at 30 °C is 42.43 mb, calculate the relative humidity.

• Actual vapor pressure (e) = Saturation vapor pressure at dew point temperature
• Saturation vapor pressure at T (es) = 6.11 * 10 ^ (7.5 * T / (237.3 + T))
• Relative Humidity (RH) = (e / es) * 100

Solution:

1. Actual Vapor Pressure:

• The dew point temperature is 20 °C. We’ll use this to calculate the actual vapor pressure.
• e = 6.11 * 10 ^ (7.5 * 20 / (237.3 + 20))
• e = 6.11 * 10 ^ (150 / 257.3)
• e = 6.11 * 10 ^ 0.583
• e ≈ 23.37 mb

2. Relative Humidity:

• The saturation vapor pressure at 30 °C is given as 42.43 mb.
• We have the actual vapor pressure (e) as 23.37 mb.
• RH = (e / es) * 100
• RH = (23.37 mb / 42.43 mb) * 100
• RH ≈ 55.07%

Answer:

1. The actual vapor pressure is approximately 23.37 mb.
2. The relative humidity is approximately 55.07%.

3. A hiker climbs a mountain, starting at an altitude of 500 meters above sea level, where the atmospheric pressure is measured to be 950 hPa (hectopascals). At the peak, the hiker measures the atmospheric pressure to be 800 hPa. Assuming a standard atmospheric lapse rate of -6.5 °C per kilometer, estimate the altitude of the peak.

The barometric formula relates pressure and altitude:

P = P₀ * exp(-Mgh/RT)

• P = Pressure at altitude h
• P₀ = Pressure at reference level (sea level, in this case)
• M = Molar mass of air (approximately 0.02896 kg/mol)
• g = Acceleration due to gravity (approximately 9.81 m/s²)
• h = Altitude
• R = Universal gas constant (8.314 J/(mol·K))
• T = Temperature (in Kelvin)

Relationship between temperature and altitude for the standard lapse rate:

T = T₀ + lapse_rate * h

Where:

• T = Temperature at altitude h
• T₀ = Temperature at reference level
• lapse_rate = -6.5 °C/km (or -0.0065 K/m)

1. Convert pressures to Pascals:

• 950 hPa = 95000 Pa
• 800 hPa = 80000 Pa

2. Assume the sea level temperature (T₀) is 20 °C (293 K). We’ll calculate the temperature at the starting altitude and then average it with the sea level temperature to get an approximate average temperature.

• T_start = T₀ + lapse_rate * h_start
• T_start = 293.15 K + (-0.0065 K/m) * 500 m
• T_start ≈ 290 K
• T_avg = (T₀ + T_start) / 2
• T_avg = (293.15 K + 290 K) / 2
• T_avg ≈ 291.57 K

3. Now, we can use the barometric formula to solve for the altitude at the summit (h).

h = -(RT/Mg) * ln(P/P₀)

• h = -((8.314 J/(mol·K)) * (291.57 K)) / ((0.02896 kg/mol) * (9.81 m/s²)) * ln(80000 Pa / 95000 Pa)
• h ≈ 1429 meters (relative to the starting altitude of 500 m)

4. Finally, we add the starting altitude to find the total altitude of the summit:

• Total altitude = 1429 m + 500 m
• Total altitude = 1929 m

Answer:

The estimated altitude of the summit is approximately 1929 meters above sea level.

4. In a certain region, the temperature at ground level is 25 °C. The normal lapse rate in this area is 6 °C per kilometer. However, a temperature inversion exists from an altitude of 1 km to 2 km, where the temperature increases at a rate of 2 °C per kilometer. 1. Calculate the temperature at an altitude of 1.5 km. 2. If a weather balloon rises from the ground to an altitude of 3 km, what temperature will it experience at 3 km?

Solution:

1. Temperature at 1.5 km:

• The first kilometer follows the normal lapse rate.
• Temperature at 1 km = 25 °C – (6 °C/km * 1 km) = 19 °C
• The next 0.5 km lies within the inversion layer.
• Temperature at 1.5 km = 19 °C + (2 °C/km * 0.5 km) = 20 °C

2. Temperature at 3 km:

• From 1.5 km to 2 km, the inversion continues.
• Temperature at 2 km = 20 °C + (2 °C/km * 0.5 km) = 21 °C
• From 2 km to 3 km, the normal lapse rate resumes.
• Temperature at 3 km = 21 °C – (6 °C/km * 1 km) = 15 °C

1. The temperature at an altitude of 1.5 km is 20 °C
2. The weather balloon will experience a temperature of 15 °C at 3 km.