Question

Air in the cylinder of a diesel engine at 20°C is
compressed from its initial pressure of 1 atm, and volume
of 200 cm to a volume of 15 cm. Assuming that air
behaves as an ideal gas (y = 1.4) and that the
compression is adiabatic, find the final pressure and
temperature.​

Answer 1

✪ANSWER✪

• $\boxed{Final Pressure = 37.58\:\:atm}$

• $\boxed{Final Temperature = 553.09⁰C}$

★EXPLAINATION★

☆GIVEN☆

• Air which behaves as Ideal gas is at 1 atm, 20⁰C in a 200 cm³ volume in a diesel engine.

• It is adiabatically compressed to 15 cm³.
• For air, $\gamma = 1.4$

☆TO FIND☆

• Final Pressure and Temperature.

᯽EQUATIONS᯽

1. If n moles of an Ideal Gas occupy volume,V at Pressure P and Temperature T, then Ideal Gas Equation gives

$\boxed{\:\:\:\:PV = nRT\:\:\:\:}$

where R is the Universal Gas Constant.

2. If an Ideal Gas undergoes adiabatic process from (P₁ , V₁) to (P₂ , V₂), then

$\boxed{\:\:\:\:P₁{V₁}^{\gamma } = P₂{V₂}^{\gamma }\:\:\:\:}$

☆CALCULATION☆

Here

$\red{\boxed{P₁ = 1 atm, V₁ = 200 cm³, T₁ = 20⁰C}}$

$\green{\boxed{P₂ = ?\:\:, V₂ = 15 cm³,T₂ =? \:\:}}$

✯ғɪɴᴀʟ ᴘʀᴇssᴜʀᴇ✯

Using above equation

$P₁{V₁}^{\gamma } = P₂{V₂}^{\gamma }$

$⇒ 1\times {200}^{1.4} = P₂{\times 15}^{1.4 }$

$⇒ 1\times {\frac{200}{15}}^{1.4} = P₂$

$⇒ P₂ = 37.58\:\:atm$

✯ғɪɴᴀʟ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ✯

If we write the the Ideal Gas Equation for state 1 and 2

$P₁V₁ = nRT₁$

$P₂V₂ = nRT₂$

Dividing, we get

$\frac{P₁V₁}{P₂V₂}=\frac{T₁}{T₂}$

$⇒\frac{1\times 200}{37.58\times 15}=\frac{(20 + 273.15)K}{T₂}$

$⇒T₂ = \frac{37.58\times 15}{1\times 200}\times 293.15$

$⇒T₂ = 826.24 K$

$⇒T₂ = 553.09⁰C$

Answer 2

compressed from its initial pressure of 1 atm, and volumeof 200 cm to a volume of 15 cm. Assuming that airbehaves as an ideal gas (y = 1.4) and that thecompression is adiabatic, find the final pressure andtemperature