Question

Air in the cylinder of a combustion engine at 20 0C is compressed from an initial pressure of 1 atmosphere and volume 8 X 10-4 m3 to a volume 6 X 10-5 m3. Assume that air behaves an ideal gas (γ = 1.40) and the process is adiabatic, calculate the final pressure.

Answer 1

Given :

• Initial pressure , P₁ = 1 atm
• Initial volume , V₁ = 8×10⁻⁴ m³
• Final volume , V₂ = 6×10⁻⁵ m³
• and γ = 1.40

To Find :

Final pressure

Theory :

For Adiabatic process :

$\bf{PV^{\bf\gamma}=constant}$

Solution :

We have to find the final pressure of gas and it is given that the process is Adiabatic.

We know in Adiabatic process :

$\sf{PV^{\gamma}=constant}$

Thus ,

$\sf{P_1\:(V_1)^{\gamma}=P_2\:(V_2)^{\gamma}}</p><p>$

Now put given the values

$\implies\sf{1\times(8\times10^{-4})^{\gamma}=P_2\:\times(6\times10^{-5})^{\gamma}}$

$\implies\sf\:P_2=(\frac{8\times10^{-4}}{6\times10^{-5}})^{1.4}$

$\implies\sf\:P_2=(\frac{80}{6})^{1.4}$

$\implies\sf\:P_2=(13.33)^{1.4}$

$\implies\sf\:P_2=37.56\:atm$

Therefore, The final pressure of the gas is 37.56 atm .

Answer 2

We have to find the final pressure of gas and it is given that the process is Adiabatic.

We know in Adiabatic process :

$PV^{y}$ = constant

We are given ,

• Initial pressure , P₁ = 1 atm

• Initial volume , V₁ = 8×10⁻⁴ m³

• Final volume , V₂ = 6×10⁻⁵ m³

• γ = 1.40

Inserting the values in this equation , the final pressure would be calculated by the given formula above .