Please help.

$Q$ amount of heat is given to $0.5$ mole of an ideal mono atomic gas by processes $TV^{n}=k$

Following graph shows variation of temperature with $Q$ . Find the value of $n$ .

#Mechanics

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Comments

We know that $dQ = nC dT$ . Since the amount of heat exchanged and temperature are both state functions and depend only on the process, we can also say that $\Delta Q = nC \Delta T$ . Thus

$C = \dfrac{1}{n} \dfrac{\Delta Q}{\Delta T} = \dfrac{1}{0.5} \dfrac{15000 - 0}{227 - 27} = 150 \text{J/K}$

If the gas undergoes the process $PV^{x} = k$ , then heat capacity for the process is given by

$C = R \left( \dfrac{1}{\gamma -1} + \dfrac{1}{1 - x} \right)$

For $TV^n = k$ , we have $PV^{n+1} = k$ and thus

$150 = R \left( \dfrac{1}{\frac 53 -1} + \dfrac{1}{-n} \right) \\ \implies \dfrac{1}{n} = \dfrac{3}{2} - \dfrac{150}{R}$

What is the unit of heat supplied?

Tapas Mazumdar - 3 years, 2 months ago

Yes, you are right $R=\frac{25}{3}$ which yields correct result. Thanks :)

Sahil Silare - 3 years, 2 months ago

@Tapas Mazumdar But a doubt still persists, that when the process is coming out to be in the form of $PV^{x}=k$ isn't it adiabatic?

Sahil Silare - 3 years, 2 months ago

@Sahil Silare – Source: Wikipedia

As you can see from the above table only when $x = \gamma$ , the process is adiabatic. In general any process of this from is a polytropic process*.

Tapas Mazumdar - 3 years, 2 months ago

@Tapas Mazumdar – Okay, have to be careful next time.

Sahil Silare - 3 years, 2 months ago

@Sahil Silare – I'm glad your concepts got clarified. :D

Tapas Mazumdar - 3 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$