Question

A diatomic gas expands adiabatically so that final density is 32 times the initial density. final pressure becomes N times of initial pressure. the value of N is?
128
1/32
32
1/128​

Answer 1

Answer:

Volume of the gas v=

density

mass

=

ρ

m

and using (PV)

r

= constant,

P

P

′

=

V

′

V

=(

mρ

mρ

)

r

=(

ρ

ρ

′

)

r

Or 128=(32)

r

∴r=log

32

128

∴r=

5

7

=1.4

I hope it helps you

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Answer 2

ANSWER.

The value of N is = 128.

EXPLANATION.

$\sf \to \: a \: diatomic \: gas \: expands \: adiabatically \\ \\ \sf \to \: final \: density \: is \: 32times \: the \: initial \: density \\ \\ \sf \to \: final \: pressure \: becomes \: n \: times \: of \: initial \: pressure.$

$\sf \to \: for \: diatomic \: gas \: = \gamma \: = \dfrac{5}{2} \\ \\ \sf \to \: as \: we \: know \: that \\ \\ \sf \to \: p_{1} v_{1} {}^{ \gamma} = p_{2} v_{2} {}^{ \gamma} = constant \\ \\ \sf \to \: \frac{ p_{1}}{ p_{2} } = ( \frac{ v_{1} }{ v_{2} } ) {}^{ \gamma}$

$\sf \to \: as \: we \: know \: that \\ \\ \sf \to \: density \: = \dfrac{mass}{volume} \\ \\ \sf \to \: d_{1} \: = \dfrac{m}{ v_{1} } \: ......(1) \\ \\ \sf \to \: d_{2} = \dfrac{m}{ v_{2} } .....(2) \\ \\ \sf \to \: from \: equation \: (1) \: and \:(2) \: we \: get$

$\sf \to \: \dfrac{ d_{1}}{ d_{2} } = \dfrac{ \dfrac{ \cancel{m}}{ v_{1} } }{ \dfrac{ \cancel{m}}{ v_{2} } } \\ \\ \sf \to \: \dfrac{ d_{1} }{ d_{2} } = \dfrac{ v_{2} }{ v_{1} } \\ \\ \sf \to \: \dfrac{ v_{1} }{ v_{2} } = \dfrac{ d_{2} }{ d_{1} } = 32 \: \: (given)$

$\sf \to \: \dfrac{ p_{2} }{ p_{1} } = (32) {}^{ \dfrac{7}{5} } \\ \\ \sf \to \: \frac{ p_{2} }{ p_{1}} = (2) {}^{ \cancel{5 \times }\dfrac{7}{ \cancel{5} }} \\ \\ \sf \to \: \frac{ p_{2} }{ p_{1} } = (2) {}^{7} = 128 \\ \\ \sf \to \: p_{2} = 128 \times p_{1} \\ \\ \sf \to \: value \: of \: n \: = 128$