Question

A sample of an ideal gas (γ = 1.5) is compressed adiabatically from a volume of 150 cm³ to 50 cm³. The initial pressure and the initial temperature are 150 kPa and 300 K. Find (a) the number of moles of the gas in the sample (b) the molar heat capacity at constant volume (c) the final pressure and temperature (d) the work done by the gas in the process and (e) the change in internal energy of the gas.

Answer 1

The ideal gas equation is
PV = nRT
Given, P₁ = 150 kPa = 150 × 10³ Pa

V₁ = 150 cm³ = 150 × 10⁻⁶ m³

T₁= 300 K

(a)

`"n" =("P""V")/("R""T") = 9.036 xx 10^-3`

n = 0.009

(b) 

`"C"_"p"/"C"_"v" = gamma , "C"_"p" -"C"_"v" ="R"`

So, `"C"_"v" = "R"/(gamma-1) = 2"R" = 8.3/0.5 = 16.6 "J" "mol" -"K"`

(c) Given,
P₁ = 150 kPa = 150 × 10³ Pa
P₂ = ? V₁ = 150 cm³
= 150 × 10⁻⁶ m³
γ = 1.5
V₂ = 50 cm³ = 50 × 10⁻⁶ m³,
T₁ = 300 K
T₂ = ?
Since the process is adiabatic, using the equation of an adiabatic process,we get
P₁V₁^γ = P₂V₂^γ
⇒ 150 × 10³ × (150 × 10⁻⁶)^γ = P₂ × (50 × 10⁻⁶)^γ

`=> "P"_2 = 150 xx 10^3 xx ((150 xx 10^-6)^1.5)/(50 xx 10 ^-6)^1.5`

P₂ = 150000 × (3)^1.5
P₂ = 779.422 × 10³ Pa
P₂ = 780 kPa
Again,
P₁^1−γ T₁^γ = P₁^1−γ T₂^γ
⇒ (150 × 10³)^1−1.5 × (330)^1.5 = (780 × 10³)^1−1.5 × T₂^1.5
⇒ T₂^1.5 = (150 × 10³)^1−1.5 × (300)^1.5 × 300^1.5
T₂^1.5 = 11849.050
⇒ T₂ = (11849.050)^1/1.5
   T₂ = 519.74 = 520 K

d) dQ = dW + dU
Or dW = −dU  [ Since dQ = 0 in an adiabatic process]
dW = −nC_vdT
dW = −0.009 × 16.6 × (520 − 300)
dW = −0.009 × 16.6 × 220
dW = −32.87 J ≈ −33 J

(e)
dU = nC_vdT
dU = 0.009 × 16.6 × 220 ≈ 33 J