JEE Mains 2020 Model Paper 3 New Pattern

pin = p0+

4T

r

⇒ ∴ pa = p0+

4T

a

, pb = p0+

4T

b

, pc = p0+

4T

c

... (i)
⇒ and Va =

4

3

πa3 , Vb =

4

3

πb3 , Vc =

4

3

πc3 ... (ii)
pV = µRT ⇒ µ =

pV

RT

; as mass is conserved µa+µb = µc
i.e.,

paVa

RTa

+

bVb

RTb

=

pcVc

RTc

⇒ paVa+pbVb = pcVc ... (iii) [Since temperature is constant]
Using (i) and (ii) in (iii)
(p0+

4T

a

)

4

3

πa3+(p0+

4T

b

)

4

3

πb3
= (p0+

t=4T

c

)

4

3

πc3
Or, 4T(a2+b2−c2) = p0(c3−a3−b3)
⇒ T =

p0(c3−a3−b3)

4(a2+b2−c2)