Question

${\LARGE{\underline{\underline{\bf{Question :-\ \textless \ br /\ \textgreater \ }}}}}$
Study the given graph (in attachement) and Identify the correct relation between $\sf{T_1}$ , $\sf{T_2}$ and $\sf{T_3}$

1] $\sf{T_1 >T_2> T_3}$

2] $\sf{T_1< T_2 < T_3}$

3] $\sf{T_1 = T_2 = T_3}$

4] $\sf{T_3> T_2 < T_1}$

Please answer with proper explanation. Spam answers will be deleted.​

Answer 1

Given three isotherms of constant temperatures $\displaystyle\sf {T_1,\ T_2}$ and $\displaystyle\sf {T_3}$ each, among which we need to find the correct relation.

Since isotherms are given, Boyle's Law is applied.

$\displaystyle\sf{\longrightarrow PV =K\quad\quad\dots(1)}$

By ideal gas equation,

$\displaystyle\sf{\longrightarrow PV=nRT\quad\quad\dots(2)}$

From (1) and (2) we get,

$\displaystyle\sf{\longrightarrow K=nRT}$

Then the work done during the process,

$\displaystyle\sf{\longrightarrow W=\int\limits_{V_1}^{V_2}P\ dV}$

$\displaystyle\sf{\longrightarrow W=\int\limits_{V_1}^{V_2}KV^{-1}\ dV}$

$\displaystyle\sf{\longrightarrow W=K\left [\log V\right]_{V_1}^{V_2}}$

$\displaystyle\sf{\longrightarrow W=nRT\log\left (\dfrac {V_2}{V_1}\right)}$

From this we get,

$\displaystyle\sf{\longrightarrow W\propto T\quad\quad\dots (3)}$

We know the work done during the process is given by area under the graph.

If $\displaystyle\sf {W_1,\ W_2}$ and $\displaystyle\sf {W_3}$ are the works done in each isotherm of temperatures $\displaystyle\sf {T_1,\ T_2}$ and $\displaystyle\sf {T_3}$ respectively, we see that,

$\displaystyle\sf{\longrightarrow W_1<W_2<W_3}$

By (3),

$\displaystyle\sf {\longrightarrow\underline {\underline {T_1<T_2<T_3}}}$

Hence (2) is the answer.

Answer 2

Answer:

Given three isotherms of constant temperatures \displaystyle\sf {T_1,\ T_2}T

1

, T

2

and \displaystyle\sf {T_3}T

3

each, among which we need to find the correct relation.

Since isotherms are given, Boyle's Law is applied.

\displaystyle\sf{\longrightarrow PV =K\quad\quad\dots(1)}⟶PV=K…(1)

By ideal gas equation,

\displaystyle\sf{\longrightarrow PV=nRT\quad\quad\dots(2)}⟶PV=nRT…(2)

From (1) and (2) we get,

\displaystyle\sf{\longrightarrow K=nRT}⟶K=nRT

Then the work done during the process,

\displaystyle\sf{\longrightarrow W=\int\limits_{V_1}^{V_2}P\ dV}⟶W=

V

1

∫

V

2

P dV

\displaystyle\sf{\longrightarrow W=\int\limits_{V_1}^{V_2}KV^{-1}\ dV}⟶W=

V

1

∫

V

2

KV

−1

dV

\displaystyle\sf{\longrightarrow W=K\left [\log V\right]_{V_1}^{V_2}}⟶W=K[logV]

V

1

V

2

\displaystyle\sf{\longrightarrow W=nRT\log\left (\dfrac {V_2}{V_1}\right)}⟶W=nRTlog(

V

1

V

2

)

From this we get,

\displaystyle\sf{\longrightarrow W\propto T\quad\quad\dots (3)}⟶W∝T…(3)

We know the work done during the process is given by area under the graph.

If \displaystyle\sf {W_1,\ W_2}W

1

, W

2

and \displaystyle\sf {W_3}W

3

are the works done in each isotherm of temperatures \displaystyle\sf {T_1,\ T_2}T

1

, T

2

and \displaystyle\sf {T_3}T

3

respectively, we see that,

\displaystyle\sf{\longrightarrow W_1 < W_2 < W_3}⟶W

1

<W

2

<W

3

By (3),

\displaystyle\sf {\longrightarrow\underline {\underline {T_1 < T_2 < T_3}}}⟶

T

1

<T

2

<T

3

Hence (2) is the answer.