Question

during the adiabatic changes video proportional to the 1/
$t {}^{2}$


how will pressure of the gass vary with the temperature ​

Answer 1

Answer:

Given:

During Adiabatic process, volume is related to temperature as follows

$V \propto \frac{1}{ {T}^{2} }$

To find:

Relationship better pressure and temperature

Concept:

We will consider Ideal Gas Equation

$PV = nRT$

Calculation:

$V \propto \frac{1}{ {T}^{2} }$

$= > \frac{nRT}{P} \propto \frac{1}{ {T}^{2} }$

Since, n and R are constants , we can say that :

$= > \frac{T}{P} \propto \frac{1}{ {T}^{2} }$

$= > P \propto \: {T}^{3}$

So final answer is :

$\boxed { \red{P \propto \: {T}^{3} }}$

Answer 2

Answer:

Taking constant "k" , we can say that:

V = k/T²

Now, putting Ideal Gas equation (PV = nRT)

=> nRT/P = k/T²

=> T/P = k"/T² .........[ k" is another constant]

=> P = (constant) T³

So P is directly proportional to T³