Answer:
a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm
Explanation:
For isothermal expansion PV = constant
So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,
So, P₁V₁ = P₂V₂
P₂ = P₁V₁/V₂
Since V₂/V₁ = 0.19,
P₂ = P₁V₁/V₂
P₂ = 1 atm (1/0.19)
P₂ = 5.26 atm
For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas
So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,
So, P₂V₂ⁿ = P₃V₃ⁿ
P₃ = P₂V₂ⁿ/V₃ⁿ
P₃ = P₂(V₂/V₃)ⁿ
Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19
1/0.19,
P₃ = P₂(V₂/V₃)ⁿ
P₃ = 5.26 atm (0.19)⁽⁵/³⁾
P₃ = 5.26 atm × 0.0628
P₃ = 0.33 atm
Using the ideal gas equation
P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)
P₃V₃/T₃ = P₄V₄/T₄
T₃ = P₃V₃T₄/P₄V₄
T₃ = (P₃/P₄)(V₃/V₄)T₂
Since V₃ = V₄ = V₁ and P₄ = P₁
V₃/V₄ = 1 and P₃/P₄ = P₃/P₁
T₃ = (P₃/P₁)(V₃/V₄)T₂
T₃ = (0.33 atm/1 atm)(1)273 K
T₃ = 90.1 K
So,
a. The highest temperature attained by the gas is T₁ = 273 K
b. The lowest temperature attained by the gas = T₃ = 90.1 K
c. The highest pressure attained by the gas is P₂ = 5.26 atm
d. The lowest pressure attained by the gas is P₃ = 0.33 atm
where s=1800 m is the runway length. Thus 
