- #1
Master1022
- 611
- 117
- Homework Statement
- Air undergoes a polytropic process from 1.2 bar and 300 K to 4 bar and 500 K. Find the polytropic exponent, n (there is more that follows on from this, but I am not interested in those bits).
- Relevant Equations
- [itex] p v^n = constant [/itex]
Hi,
I was doing this question and I was slightly confused as to whether I ought to just substitute [itex] n = \gamma [/itex] (the adiabatic constant) into the equation? The answers don't do this, but I was wondering why it was wrong for me to do so? This was only a small fraction of the question (which was worth very few marks), so I thought it would be an appropriate substitution to make given that we often assume air is a perfect gas.
I cannot really understand the reason not to use [itex] n = \gamma [/itex], apart from it yielding the wrong answer (correct answer is ~1.74 and I just let n = 1.4 from our textbook data table).
The answer scheme seems to suggest using: [tex] p_{1} v_1 ^ n = p_{2} v_2 ^ n [/tex]
[tex] n = \frac{\log(\frac{p_2}{p_1})}{\log(\frac{v_1}{v_2})} [/tex]
and we can calculate v1 and v2 from the conditions given.
Thanks in advance
I was doing this question and I was slightly confused as to whether I ought to just substitute [itex] n = \gamma [/itex] (the adiabatic constant) into the equation? The answers don't do this, but I was wondering why it was wrong for me to do so? This was only a small fraction of the question (which was worth very few marks), so I thought it would be an appropriate substitution to make given that we often assume air is a perfect gas.
I cannot really understand the reason not to use [itex] n = \gamma [/itex], apart from it yielding the wrong answer (correct answer is ~1.74 and I just let n = 1.4 from our textbook data table).
The answer scheme seems to suggest using: [tex] p_{1} v_1 ^ n = p_{2} v_2 ^ n [/tex]
[tex] n = \frac{\log(\frac{p_2}{p_1})}{\log(\frac{v_1}{v_2})} [/tex]
and we can calculate v1 and v2 from the conditions given.
Thanks in advance