MCQ
$6$ moles of an ideal gas expand isothermally and reversibly from a volume of $1$ litre to a volume of $10$ litres at $27\,^oC$. What is the maximum work done? ................. $\mathrm{kJ}$
- A$47$
- B$100$
- C$0$
- ✓$34.465$
Given $n = 6,T = 27{\,^o}C = 273 - 27 = 300\,K$
${V_1} = 1\,L,\,{V_2} = 10\,L$
$\therefore W = - 2.303 \times 6 \times 8.314 \times 300\,\log \,\frac{{10}}{1}$
$ = 34.465\,kJ$
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$(A)$ the reaction of $Al _2 O _3$ with coke ($C$) at a temperature $>2500^{\circ} C$.
$(B)$ the neutralization of aluminate solution by passing $CO _2$ gas to precipitate hydrated alumina $\left( Al _2 O _3 .3 H _2 O \right)$
$(C)$ the dissolution of $Al _2 O _3$ in hot aqueous $NaOH$.
$(D)$ the electrolysis of $Al _2 O _3$ mixed with $Na _3 AlF _6$ to give $Al$ and $CO _2$.
$A. \;n =3, l=2, m _{1}=1, m _{ s }=+1 / 2$
$B.\; n =4, l=1, m _{1}=0, m _{ s }=+1 / 2$
$C. \;n =4, l=2, m _{1}=-2, m _{ s }=-1 / 2$
$D. \;n =3, l=1, m _{1}=-1, m _{ s }=+1 / 2$
The correct order of increasing energy is
$(A)$ $(a)$ $n=3,1=1, m_{l}=1, m_{s}=+\frac{1}{2}$
$\quad (b)$ $n =3,1=2, m _{l}=1, m _{s}=+\frac{1}{2}$
$(B)$ $(a)$ $n =3,1=2, m _{l}=-2, m _{s}=-\frac{1}{2}$
$\quad (b)$ $n =3,1=2, m _{l}=-1, m _{s}=-\frac{1}{2}$
$(C)$ $(a)$ $n=4,1=2, m_{l}=2, m_{s}=+\frac{1}{2}$
$\quad (b)$ $n =3,1=2, m _{l}=2, m _{s}=+\frac{1}{2}$
The pairs of electron present in degenerate orbitals is/are...... .