First Law of Thermodynamics

Dec 2018
13
1
The 'Milky Way' Galaxy
I read from a famous text-book :

"when mechanical work is done on a system of gas, the internal energy of the system increases."

According to the formula of the First Law of Thermodynamics,it sounds true.
But ,theoretically, when we do work on a gas , it compresses,i.e.,its volume decreases.When volume decreases than temperature should also decrease as 'v' is directly proportional to 't'. In case temperature decreases, than internal energy will also decrease.
This is a clear-cut contradiction of the above statement.

I feel i am wrong somewhere because the statement can't be wrong.

Plz somebody help me point out my mistake.


Thanks in advance!
 
Oct 2017
660
331
Glasgow
But ,theoretically, when we do work on a gas , it compresses,i.e.,its volume decreases.
Yes...

When volume decreases than temperature should also decrease as 'v' is directly proportional to 't'.
This is not necessarily the case as it depends on the pressure. State transitions in an ideal gas vary as

\(\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

So, at constant pressure (\(\displaystyle P_1 = P_2\)), temperature increases if the volume decreases. This is usually what happens under a compression.

EDIT: ooops, I mean "at constant pressure, temperature decreases if volume decreases"
 
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Dec 2018
13
1
The 'Milky Way' Galaxy
Yes...



This is not necessarily the case as it depends on the pressure. State transitions in an ideal gas vary as

\(\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

So, at constant pressure (\(\displaystyle P_1 = P_2\)), temperature increases if the volume decreases. This is usually what happens under a compression.
How do you say this , I really don't understand and believe it is incorrect because volume is directly proportional to temperature.
PV=nRT
 
Oct 2017
660
331
Glasgow
volume is directly proportional to temperature.
Oooops... I made a mistake. Sorry, here's a better explanation:

In isobaric compression, heat is transferred to or from the system to ensure that the pressure remains constant as the gas performs work or has work done on it

i) \(\displaystyle P_2 = P_1\) and;
ii) \(\displaystyle \frac{V_1}{T_1} = \frac{V_2}{T_2}\)

So... if you decrease the volume (\(\displaystyle V_2 < V_1\)), then temperature decreases.


However, if we instead consider "Adiabatic compression", where no heat is transferred to or from the system (Q = 0), the formula describing adiabatic compression is not given by the ideal gas law (even for an ideal gas). It is given by:

\(\displaystyle TV^{\gamma-1} = \text{constant}.\)

where \(\displaystyle \gamma\) is the ratio of heat capacities and is 1.66 for an ideal gas. We can do a similar set of mathematics above to obtain

\(\displaystyle T_1 V_1^{0.66} = T_2 V_2^{0.66}\)

So we get a different behaviour... because \(\displaystyle V_2\) is smaller than \(\displaystyle V_1\), \(\displaystyle T_2\) must increase.

So... the impact on the state of the gas depends on the kind of transition that is being done because there are three state variables in the formula, not two. You can use the ideal gas law to determine what the corresponding pressures are for those two states.
 
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topsquark

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How do you say this , I really don't understand and believe it is incorrect because volume is directly proportional to temperature.
PV=nRT
Just for the record, for an ideal gas:
\(\displaystyle PV = nRT\)

So for two states 1 and 2 we have
\(\displaystyle P_1 V_1 = nRT_1 \implies nR = \dfrac{P_1V_1}{T_1}\)

\(\displaystyle P_2 V_2 = nRT_2 \implies nR = \dfrac{P_2V_2}{T_2}\)

So
\(\displaystyle nR = \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\)

just as benet13 was saying.

Of course that was not what you needed here but you really need to have this equation in your back pocket.

-Dan