The temperature of the earth’s atmosphere drops about 5℃ for every 1000m of elevation above earth's surface. The earth temperature at ground level is 11.33℃ and the pressure is 760 mmHg. Solve for the elevation at which the pressure is 380 mmHg. Assume that Air is an ideal gas
To solve for the elevation at which the pressure is 380 mmHg, we can use the barometric formula, which relates the pressure and temperature changes with altitude for an ideal gas.
The barometric formula is expressed as:
P2/P1 = (T2/T1)^((g*M)/(R*L))
Where:
P1 and T1 are the initial pressure and temperature at the ground level
P2 and T2 are the pressure and temperature at the desired elevation
g is the acceleration due to gravity (approximately 9.81 m/s^2)
M is the molar mass of air (approximately 0.02896 kg/mol)
R is the ideal gas constant (approximately 8.314 J/(mol·K))
L is the temperature lapse rate, which represents the rate at which temperature changes with elevation (given as 5°C per 1000m in this case)
Let's plug in the values and solve for the desired elevation (h):
P2/P1 = (T2/T1)^((g*M)/(R*L))
P2/760 mmHg = (T2/(11.33°C+273.15 K))^((9.81 m/s^2 * 0.02896 kg/mol)/(8.314 J/(mol·K) * -5°C per 1000m))
Simplifying:
P2/760 = (T2/284.48)^(-0.03588)
To solve for T2, we can rearrange the equation:
T2 = 284.48 * (P2/760)^(-27.829)
Let's calculate T2 using the given pressure (380 mmHg):
T2 = 284.48 * (380/760)^(-27.829)
T2 ≈ 284.48 * (0.5)^(-27.829)
T2 ≈ 284.48 * 3.007
T2 ≈ 854.24 K
Now that we have T2, we can calculate the elevation (h) using the following equation:
h = (T2 - T1) / L
h = (854.24 K - 11.33°C) / 5°C per 1000m
Converting 11.33°C to Kelvin:
h ≈ (854.24 K - 284.48 K) / 5°C per 1000m
h ≈ 569.76 K / 5°C per 1000m
Converting to meters:
h ≈ 113.952 m / 1000m per km
Therefore, the elevation at which the pressure is 380 mmHg is approximately 0.114 km.
To solve for the elevation at which the pressure is 380 mmHg, we can use the barometric formula, which relates pressure, temperature, and elevation. The barometric formula is given by:
P = P0 * exp(-(M * g * h) / (R * T))
Where:
- P is the pressure at a given elevation
- P0 is the pressure at ground level
- M is the molar mass of air
- g is the acceleration due to gravity
- h is the elevation
- R is the ideal gas constant
- T is the temperature in Kelvin
Given:
- P0 = 760 mmHg
- T0 = 11.33℃ (convert to Kelvin by adding 273.15: T0 = 11.33 + 273.15 = 284.48 K)
- P = 380 mmHg
Assuming that the molar mass of air (M), the acceleration due to gravity (g), and the ideal gas constant (R) remain constant, we can rearrange the equation to solve for the elevation (h):
h = -(R * T * log(P / P0)) / (M * g)
Now let's calculate the elevation:
Step 1: Convert pressures to atmospheres (1 atm = 760 mmHg)
P0 = 760 mmHg / 760 mmHg = 1 atm
P = 380 mmHg / 760 mmHg = 0.5 atm
Step 2: Calculate the elevation (h)
h = -(R * T * log(P / P0)) / (M * g)
= -(8.314 J/(mol*K) * 284.48 K * log(0.5)) / (0.02897 kg/mol * 9.8 m/s^2)
Calculating the value of the expression above, we can find the elevation in meters.
p = 760 - 5 * h/1000
so, now just find h when
760 - .005h = 380