Question

Topic :- Kinetic theory of gases

Show your working :)​

Answer 1

Answer:

$\large \bf\clubs \: Given :-$

$\underline{\purple{\ddot{\Mathsdude}}}$

☆♧Answered by Rohith kumar maths dude :-

♧♧Given:-

PV diagram of two different masses m1 and m2.

♧♧To prove :-

State wheather m1 >m2 or m2 >m1

♧♧Proof:-

Here By using ideal gas equation in both of the cases .

First we take,

♧♧Assumption :- both gases are identical only.

●And taking molar mass in each cases is M.

Let take formula of ideal gas equation: -

♧(i) Case:-

PV=n1RT

PV=m1 (RT/M) -----(i)

♧(ii) Case:-

PV=n2RT

PV= m2 (RT/M) ------(ii)

♧♧♧Let,

v=vnot from the graph

P1 and P2 are pressures

And P2>P1 ----(iii).

♧♧P1V not= m1(RT/M)

♧♧P1= m1/V not (RT/M)

♡♡P2V not= m2 (RT/M)

♡♡P2= m2/V not (RT/M)

♧♧Now solving ,

♧♧P2>P1

Applying the P1 and P2 values,

=m1/V not (RT/M) > m2/V not (RT/M)

♧From this we get ,

= m1 > m2

♧☆Hope it helps u @Mystified boy (Apprentice moderator)

☆☆Thank you for good question .

Answer 2

Answer:

Answer:

\large \bf\clubs \: Given :-♣Given:−

\underline{\purple{\ddot{\Mathsdude}}}

\Mathsdude

¨

☆♧Answered by Rohith kumar maths dude :-

♧♧Given:-

PV diagram of two different masses m1 and m2.

♧♧To prove :-

State wheather m1 >m2 or m2 >m1

♧♧Proof:-

Here By using ideal gas equation in both of the cases .

First we take,

♧♧Assumption :- both gases are identical only.

●And taking molar mass in each cases is M.

Let take formula of ideal gas equation: -

♧(i) Case:-

PV=n1RT

PV=m1 (RT/M) -----(i)

♧(ii) Case:-

PV=n2RT

PV= m2 (RT/M) ------(ii)

♧♧♧Let,

v=vnot from the graph

P1 and P2 are pressures

And P2>P1 ----(iii).

♧♧P1V not= m1(RT/M)

♧♧P1= m1/V not (RT/M)

♡♡P2V not= m2 (RT/M)

♡♡P2= m2/V not (RT/M)

♧♧Now solving ,

♧♧P2>P1

Applying the P1 and P2 values,

=m1/V not (RT/M) > m2/V not (RT/M)

♧From this we get ,

= m1 > m2

♧☆Hope it helps u

☆☆Thank you for good question .