To a good approximation, the temperature T drops almost linearly with altitude up to 11 km above sea level, at a constant rate:

dTdz=−α forz≤11 km

where α=6.5 K/km (Kelvin per km) and z is the height above the sea level. The temperature stays then approximately constant between 11 km and 20 km above sea level.

Assume a temperature of 5 ∘C and a pressure of 1 atm at sea level (1 atm = 1.01325 ×105 N/m^2). Furthermore, take the molecular weight of the air to be (approximately) 29 g/mol. The universal gas constant is R=8.314 JK−1mol−1 and the acceleration due to gravity is g=10 m/s2 (independent of altitude). Assume that air can be treated as an ideal gas.

(a) Under the assumptions above, calculate the atmospheric pressure p (in atm) at z= 10 km above sea level for the case of a linear temperature drop.

p=

(b) The cruising altitude of a commercial aircraft is about 33'000 ft (or 10 km). Assume that the cabin is pressurized to 0.8 atm at cruising altitude. What is the minimal force Fmin (in Newton) per square meter that the walls have to sustain for the cabin not to burst? Use the atmospheric pressure found in (a).

Fmin=

(c) We close a plastic bottle full of air inside the cabin when the aircraft is at cruising altitude of z= 10 km. The volume of the bottle is V1, the pressure and temperature inside the cabin are 0.8 atm and T1=27 ∘C, respectively. Assume that at sea level the atmospheric pressure is 1 atm, and the temperature is decreased by 15 Kelvin with respect to the cabin's temperature.

What is the magnitude of the percentage change in volume of the air inside the bottle when it is brought to sea level? (Enter the magnitude of the percentage change in volume in

∣∣∣ΔVV1∣∣∣×100=

Guys please help if you can !

(a) To calculate the atmospheric pressure at z = 10 km above sea level, we need to first find the temperature at that altitude using the given linear temperature drop equation.

dT/dz = -α

where α = 6.5 K/km is the constant rate of temperature drop up to 11 km above sea level.

Since we know the temperature at sea level is 5 °C, we convert it to Kelvin:

T0 = 5 °C + 273.15 = 278.15 K

Using the linear temperature drop equation, we can calculate the temperature at z = 10 km:

T10 = T0 + dT/dz * (10 - 0) = 278.15 K - 6.5 K/km * 10 km = 278.15 K - 65 K = 213.15 K

Now, we can use the ideal gas law to calculate the pressure at z = 10 km using the calculated temperature:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature.

Assuming the volume and number of moles of air remain constant, we can write:

P0 / T0 = P / T10

Solving for P:

P = P0 * T10 / T0

Substituting the given values:

P = (1 atm) * 213.15 K / 278.15 K = 0.766 atm

Therefore, the atmospheric pressure at z = 10 km above sea level is approximately 0.766 atm.

(b) The minimal force (Fmin) per square meter that the cabin walls have to sustain for the cabin not to burst can be calculated using the pressure difference between the inside and outside of the cabin.

The pressure difference is given by:

ΔP = Pinside - Poutside

Given that the pressure inside the cabin is 0.8 atm and the atmospheric pressure outside (from part a) is approximately 0.766 atm:

ΔP = 0.8 atm - 0.766 atm = 0.034 atm

Now, we can convert this pressure difference to force per square meter using the ideal gas law:

ΔP = F / A

where F is the force and A is the area.

Rearranging the equation, we have:

Fmin = ΔP * A

To calculate A, we need to know the area of the cabin walls. Let's assume it is 1 square meter for simplicity.

Therefore,

Fmin = 0.034 atm * 1 m² = 0.034 atm * 1.01325 × 105 N/m²/atm = 3,445 N

So, the minimal force per square meter that the walls have to sustain for the cabin not to burst is approximately 3,445 Newtons.

(c) To calculate the magnitude of the percentage change in volume of the air inside the plastic bottle when it is brought to sea level, we can use the combined gas law equation:

P1 * V1 / T1 = P2 * V2 / T2

where P1 and T1 are the pressure and temperature of the air inside the bottle at cruising altitude, and P2 and T2 are the pressure and temperature at sea level.

Given that P1 = 0.8 atm, T1 = 27 °C, P2 = 1 atm, and T2 = T1 - 15 K (decreased temperature by 15 Kelvin with respect to the cabin's temperature), and assuming V2 is the volume at sea level that we need to find, we can rearrange the equation as follows:

V2 = (P1 * V1 * T2) / (T1 * P2)

Substituting the given values:

V2 = (0.8 atm * V1 * (27 °C - 15 K)) / (27 °C * 1 atm)

Simplifying:

V2 = 0.8 * V1 * (27 - 15) / 27

Now, to find the magnitude of the percentage change in volume, we can use the following formula:

|ΔV / V1| * 100

where ΔV = V2 - V1

Substituting the calculated values into the formula, we have:

|ΔV / V1| * 100 = (|V2 - V1| / V1) * 100 = (|0.8 * V1 * (27 - 15) / 27 - V1| / V1) * 100

Simplifying further, we get:

|(0.8 * V1 * 12 / 27 - V1)| / V1 * 100

Hence, the magnitude of the percentage change in volume of the air inside the bottle when it is brought to sea level is:

|(0.8 * V1 * 12 / 27 - V1)| / V1 * 100