One mole of an ideal gas at $$300\,K$$ is expanded isothermally from an initial volume of $$1\,L$$ to $$10L.$$ The $$\Delta E$$ for this process is $$\left( {R = 2\,cal\,mo{l^{ - 1}}{K^{ - 1}}} \right)$$
A.
$$163.7\,cal$$
B.
$${\text{Zero}}$$
C.
$$1381.1\,cal$$
D.
$$9\,L\,atm$$
Answer :
$${\text{Zero}}$$
Solution :
Isothermal process means temperature remains constant. At constant temperature, internal energy $$\left( {\Delta E} \right)$$ also remains constant. So, $$\Delta E = 0$$
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 $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$