Question

15 g of nitrogen gas at STP are adiabatically compressed to a pressure of 21 atm .

What is the final temperature of the gas?

What is the work done on the gas?

What is the heat input to the gas?

What is the compression ratio Vmax/Vmin?

Answer 1

1)

An ideal gas undergoing an adiabatic, reversible process satisfies the condition:

p∙V^γ = constant

throughout the process.

γ = Cp/Cv is the heat capacity ratio. For a diatomic ideal gas like nitrogen it is

γ = (7/2)∙R / (5/2)∙R = 7/5

Using ideal gas you may rewrite adiabatic equation as:

p^(γ-1)∙T^(-γ) = constant

Hence:

p₁^(γ-1) / T₁^γ = p₂^(γ-1) / T₂^γ

T₂ = T₁ ∙ (p₂/p₁)^[(γ-1)/γ]

= 273.15K ∙ (21atm / 1atm)^[(7/5-1)/(7/5)]

= 273.15K ∙ (21)^[2/7]

= 651.90 K

2)

The work done on the gas is given by the integral

W = - ∫ p dV from V₁ to V₂

V can expressed in terms of V by adiabatic equation:

p∙V^γ = constant = p₁∙V₁^γ

V = V₁∙p₁^(1/γ) ∙p^(-1/γ)

dV = V₁∙p₁^(1/γ) ∙ (-1/γ)∙p^(-1/γ - 1) dp

W = - ∫ p dV from V₁ to V₂

= - ∫ p ∙ V₁∙p₁^(1/γ) ∙ (-1/γ)∙p^(-1/γ - 1) dp from p₁ to p₂

= V₁∙p₁^(1/γ) ∙ (1/γ) ∙ ∫ p^(-1/γ) dp from p₁ to p₂

= V₁∙p₁^(1/γ) ∙ (1/γ) ∙ [1/(1 - 1/γ)] ∙ { p₂^[(1 - 1/γ)] - p₁^[(1 - 1/γ)] }

= V₁∙p₁^(1/γ) ∙ [1/(γ - 1)] ∙ { p₂^[(1 - 1/γ)] - p₁^[(1 - 1/γ)] }

= V₁∙p₁ ∙ [1/(γ - 1)] ∙ { (p₂/p₁)^[(1 - 1/γ)] - 1 }

= n∙R∙T₁ ∙ [1/(γ - 1)] ∙ { (p₂/p₁)^[(1 - 1/γ)] - 1 }

For this process

W = (15g/28g/mol) ∙ 8.314472J/molK ∙ 273.15K ∙ (5/2) ∙ { (21)^(2/7) -1}

W = 4217.27J

4)

T1 = 273.15 K

T2 = 651.9 K

Gas ideal equation,

PV = nRT

PV/T = nR = constant

P1V1/T1 = P2V2/T2

V2/V1 = (P1/P2) * (T2/T1) .... (1)

PV^k = constant

P1V1^k = P2V2^k

P1/P2 = (V2/V1)^k ............. (2)

(1) --> (2)

V2/V1 = (V2/V1)^k * (T2/T1)

(V2/V1)^1-k = (T2/T1)

(V2/V1) = (T2/T1)^1/(1-k)

V2 / V1 = (651.9/273.15)^(1/1-1.4)

V2 / V1 = 0.1136

V_max/Vmin = 8.799