Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

1.77

Explanation:

We know that ,in isothermal process

$w1\: = nrt ln( \frac{v2}{v1} )$

Already given that , initial pressure p1 and V1 ,after expantion volume becomes 4v1

Since,

$w 1\: = nrt \: ln( \frac{4v1}{v1} ) = 2nrt \: log2$

$\blue{in \adiabatic \: process}$

After isothermal process goes into adiabatic .

then changes happened with volume volume becomes 4v1 to 32v2

we know a relation in adiabatic process .

i.e

$T1(v1) {}^{ \gamma - 1} = T2(v2) {}^{ \gamma - 1}$

From this relation.

$= > t1(4v1) {}^{ \gamma - 1} = t2(32v1) {}^{ \gamma - 1}$

$= > t2 =t 1(4v1) {}^{ 1 - \gamma}$

In mono atomic gas ,

$\gamma = \frac{5}{3}$

$t2 = t1(8v1) ] {}^{ 1 - \frac{5}{3} }$

$t2 = t1(8v1) {}^{ \frac{2}{3} } = \frac{t1}{4}$

W2 = - ncvmR(∆T) = ncvmR (t2-t1)

In monoatimic cvm = 3/2

=> W2 = 9/8 n rt

Wiso/Wadia = ?

$\frac{w1}{w2} = \frac{2nrt \: ln2}{ \frac{9}{8}nrt } = \frac{16}{9} ln2$

Since, 16/9 = 1.77 Answer

Answer 2

ANSWER.

The value of f = 1.78

EXPLANATION.

$\sf : \implies \: p_{1} v_{1} \: \longrightarrow \: p_{1} 4v_{1} \\ \\ \sf : \implies \: w_{isothermally} \: = NRT \: ln( \frac{ v_{2} }{ v_{1} } ) \\ \\ \sf : \implies \: w_{isothermally} \: = NRT \: ln( \frac{4 v_{1}}{ v_{1} } ) \: \: \: .....(1)$

$\sf : \implies \: 4 v_{1} \: \longrightarrow \: 32 v_{1} \\ \\ \sf : \implies \: as \: we \: know \: that \\ \\ \sf : \implies \: t_{1} v_{1} {}^{ \gamma - 1} = t _{2} v_{2} {}^{ \gamma - 1} \\ \\\sf : \implies \: t(4v) {}^{ \gamma - 1} = t_{2}(32v) {}^{ \gamma - 1} \\ \\ \sf : \implies \: t_{2} \: = t \: 8 {}^{( \dfrac{ - 2}{3}) }$

$\sf : \implies \: w_{adiabatically} \: = \dfrac{nr \Delta \: t}{1 - \gamma } = \dfrac{nr( t_{2} - t) }{1 - \frac{5}{3} } \\ \\ \sf : \implies \: helium \: is \: a \: monoatomic \: gas \: = \gamma \: = \frac{5}{3} \\ \\ \sf : \implies \: w_{adiabatically} \: = \frac{nr \: (8 {}^{ \dfrac{ - 2}{3} } - 1) t}{ \dfrac{ - 2}{3} } \\ \\ \sf : \implies \: w_{adiabatically} \: = \frac{9}{8}nrt \: \: \: .....(2)$

$\sf : \implies \: from \: equation \: (1) \: \: and \: \: (2) \: \: we \: get \: \\ \\ \sf : \implies \: \frac{ w_{iso} }{ w_{adia}} = \frac{2 ln(2) }{ \frac{9}{8} } = \frac{16}{9} ln(2) = 1.78$