Air is compressd in a cylinder

[Karol]

Homework Statement

At the start of the compression stroke, the cylinder in a diesel motor contains 300[cm³] of air at atmospheric pressure and 27⁰C, and at the end of the stroke the volume is 20[cm³] and manometric pressure 41[atm]. What is the temperature.

Homework Equations

$$PV=nRT\rightarrow \frac{P_1V_1}{T_1}={P_1V_2}{T_2}$$
In an adiabatic process: $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$
And: $P_1V_1^{\gamma}=P_2V_2^{\gamma}$
γ for air=1.4

The Attempt at a Solution

First i check if this is an adiabatic process by comparing the initial and final volumes and pressures:
$$T_1V_1^{\gamma-1}=1\cdot 300^{1.4}=2937$$
$$T_2V_2^{\gamma-1}=42\cdot 20^{1.4}=2784$$
So it's not, heat leaks.
So i calculate using the equation of state:
$$\frac{P_1V_1}{T_1}={P_1V_2}{T_2}\rightarrow \frac{1\cdot 300}{300}=\frac{42\cdot 20}{T_2}\rightarrow T_2=840^0K$$
Is my method true?

 

[Chestermiller]

Yes. Note that this is strictly an ideal gas problem. You didn't need to use the equations with the gammas. But, if you had, would your results have been consistent with the gamma equations?

Chet

 

[Karol]

But, if you had, would your results have been consistent with the gamma equations?

You mean if i solved the gamma equations? i guess not since it isn't adiabatic, i don't understand the question, i think

 

[Chestermiller]

You mean if i solved the gamma equations? i guess not since it isn't adiabatic

Yes, that's what I meant, and your conclusion is correct.

Chet

 

[HalfLostAndSearching]

Yes, your method is correct. By using the ideal gas law and assuming that the process is isothermal, you were able to calculate the final temperature of the air in the cylinder. However, as you noted, this is not an adiabatic process as heat is lost during compression. If you wanted to take into account the heat loss, you could use the adiabatic equations and solve for the final temperature using the initial temperature and volume, along with the final pressure and volume. This would give you a more accurate result, but your approach is still valid for an approximate calculation.