Study the given graph (in attachement) and Identify the correct relati

1.	Study the given graph (in attachement) and Identify the correct relation between , and 1] 2] 3] 4] Please answer with proper explanation. Spam answers will be deleted.
Answer» 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, $\displaystyle\sf{\longrightarrow W_1$ By (3), $\displaystyle\sf {\longrightarrow\underline {\underline {T_1$ Hence (2) is the answer.

Study the given graph (in attachement) and Identify the correct relation between , and 1] 2] 3] 4] Please answer with proper explanation. Spam answers will be deleted.

Answer»

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,