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Basic thermodynamics

Air at 5 bar, 500K is contained in a cylinder by a piston, and initially occupies a volume of 1m^3. 100kJ of heat are transferred to the air, whilst the volume is kept constant. The temperature in the cylinder after this is 540k. The mass of the air is 3.48kg.

Calculate the pressure in the cylinder at the end of the process.

Using the ideal gas law...

P2=mRT2V=3.48⋅287⋅5401=5.4⋅105P_2 = \frac {mRT_2}{V} = \frac {3.48 \cdot 287 \cdot 540}{1} = 5.4 \cdot 10^5

Using isentropic relations...

P2=P1⋅(T2T1)γγ−1=5⋅105⋅(540500)1.41.4−1=6.5⋅105P_2 = P_1 \cdot (\frac {T_2}{T_1})^{\frac {\gamma}{\gamma -1}} = 5 \cdot 10^5 \cdot (\frac {540}{500})^{\frac {1.4}{1.4-1}} = 6.5 \cdot 10^5

Why do I get a different answer? I'm pretty sure that both methods should work..correctly that is
Original post by Serendreamers
Air at 5 bar, 500K is contained in a cylinder by a piston, and initially occupies a volume of 1m^3. 100kJ of heat are transferred to the air, whilst the volume is kept constant. The temperature in the cylinder after this is 540k. The mass of the air is 3.48kg.

Calculate the pressure in the cylinder at the end of the process.

Using the ideal gas law...

P2=mRT2V=3.48⋅287⋅5401=5.4⋅105P_2 = \frac {mRT_2}{V} = \frac {3.48 \cdot 287 \cdot 540}{1} = 5.4 \cdot 10^5

Using isentropic relations...

P2=P1⋅(T2T1)γγ−1=5⋅105⋅(540500)1.41.4−1=6.5⋅105P_2 = P_1 \cdot (\frac {T_2}{T_1})^{\frac {\gamma}{\gamma -1}} = 5 \cdot 10^5 \cdot (\frac {540}{500})^{\frac {1.4}{1.4-1}} = 6.5 \cdot 10^5

Why do I get a different answer? I'm pretty sure that both methods should work..correctly that is


I'm by no means certain but the isetropic derivation probably doesn't involve perfect gases if it gives a different answer.
Reply 2
Avatar for Borek
Borek
4
Why do you think the process is isentropic?

dS=dQ/T, is dQ zero?
Wtf? Is this really sixth form stuffs?

Posted from TSR Mobile
Reply 4
Avatar for Serendreamers
Serendreamers
OP
1
Original post by Borek
Why do you think the process is isentropic?

dS=dQ/T, is dQ zero?


Ahh, I'm such a fool. Thank you.

Original post by QuantumSuicide
Wtf? Is this really sixth form stuffs?

Posted from TSR Mobile


Nope, that was an accidental mis-labeling. Please forgive me :biggrin:

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