Question

In the course of pumping up a bicycle tire, a liter of air at atmospheric pressure is compressed ISOTHERMALLY to a pressure of 7 atm. (Air is mostly diatomic and oxygen).
(a) if the temperature is initially 300 K, what is the temperature after compression?
(b) what is the final volume of this air after compression?
(c) how much work is done in compressing the air?

Answer 1

a)
Pressure and volume of an ideal gas undergoing an reversible adiabatic process satisfy the relation:
P∙V^γ = constant
The exponent γ is the heat capacity ratio:
γ = Cp/Cv

The molar heat capacities for an ideal gas with molecules having f degrees of freedom are:
Cv = (f/2)∙R
Cp = Cv + R = ((f + 2)/2)∙R
Hence,
γ = (f + 2)/f
for air
γ = (7/5)

From the adiabatic equation above follows that pressure and volume in initial state (1) and final state (2) of the process are related as:
P₁∙V₁^γ = P₂∙V₂^γ
So final volume is given by:
V₂ = V₁∙(P₁/P₂)^(1/γ)
= 1 L ∙ (1 atm / 7 atm)^(5/7)
= 0.249 L


b)
There is not heat transfer in an adiabatic process..So the change in internal energy of the gas equals the work done one the gas in the process.
∆U = W

Internal energy for an ideal gas is given by:
U = n∙Cv∙R∙T = (f/2)∙n∙R∙T
So work done in this process is:
W = (f/2)∙n∙R∙(T₂ - T₁)
Using ideal gas law you can express temperature in each state in terms of pressure and volume:
n∙R∙T₁ = P₁∙V₁
n∙R∙T₂ = P₂∙V₂
Hence,
W = (f/2)∙(P₂∙V₂ - P₁∙V₁)
=>
W = (5/2)∙(7∙101325 Pa ∙ 0.249×10⁻³ m³ - 1∙101325 Pa ∙ 1×10⁻³ m³)
= 188.4 Pa∙m³
= 188.4 J


c)
From ideal gas law follows that
P∙V/T = n∙R = constant
So
P₁∙V₁/T₁ = P₂∙V₂/T₂
=>
T₂ = T₁ ∙ (P₂/P₁) ∙ (V₂/V₁)
= 300 K ∙ (7/1) ∙ (0.249/1)
= 523 K