A given mass of gas expands from state $$A$$ to state $$B$$ by three paths 1, 2, and 3 as shown in the figure :
If $${w_1},{w_2}$$ and $${w_3}$$ respectively, be the work done by the gas along three paths, then
A.
$${w_1} > {w_2} > {w_3}$$
B.
$${w_1} < {w_2} < {w_3}$$
C.
$${w_1} = {w_2} = {w_3}$$
D.
$${w_2} > {w_1} > {w_3}$$
Answer :
$${w_1} < {w_2} < {w_3}$$
Solution :
Work is equal to area under $$P - V$$ graph ( when $$P$$ is plotted along $$y$$ - axis ). As area under graph 3 is maximum and area under graph 1 is minimum, so $${w_3}$$ is minimum and $${w_1}$$ is minimum.
$$\therefore \,\,{w_1} < {w_2} < {w_3}$$
Releted MCQ Question on Physical Chemistry >> Chemical Thermodynamics
Releted Question 1
The difference between heats of reaction at constant pressure and constant volume for the reaction : $$2{C_6}{H_6}\left( l \right) + 15{O_{2\left( g \right)}} \to $$ $$12C{O_2}\left( g \right) + 6{H_2}O\left( l \right)$$ at $${25^ \circ }C$$ in $$kJ$$ is
$${\text{The}}\,\Delta H_f^0\,{\text{for}}\,C{O_2}\left( g \right),\,CO\left( g \right)\,$$ and $${H_2}O\left( g \right)$$ are $$-393.5,$$ $$-110.5$$ and $$ - 241.8\,kJ\,mo{l^{ - 1}}$$ respectively. The standard enthalpy change ( in $$kJ$$ ) for the reaction $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$