NEET Practice Questions (MCQs) with Answers & Solutions

(a) q=$∆$U-W, if adiabatic process q=0, then -$∆$U = -W, i.e, a decrease in free energy brings in work done by the system (-W).

(a) Entropy, $∆{\mathrm{S}}_{\mathrm{f}}=∆{\mathrm{H}}_{\mathrm{f}}/{\mathrm{T}}_{\mathrm{f}}$= Fusion enthalpy/Temperature

                  $∆{\mathrm{S}}_{\mathrm{f}}$ = 2930 J mol^-1/300 K = 9.77 JK^-1mol^-1

(a) -W = +2.303nRTlog$\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

      -W = 2.303 x 16/32 x 300 x 8.314log(25/5)

      -W=2.01 x 10³ J

(a) $∆$S=q/T =$\frac{2.303\mathrm{nRT}}{\mathrm{T}}\log \frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$=2.303 x 1 x 8.314log10

                                          = 19.15 JK^-1mol^-1

(c) An adiabatic process is one in which exchange of heat is not taking place in between system and surroundings. This can be made by putting insulation at the boundaries of system.

(a) For reaction,

C(s) + 2H₂(g) $\to$CH₄(g) , $∆\mathrm{H}°$=?

 $∆\mathrm{H}°$=-[$∆{\mathrm{H}}_{\mathrm{c}}^{°}$ of CH₄) -($∆{\mathrm{H}}_{\mathrm{c}}^{°}$of C +2x $∆{\mathrm{H}}_{\mathrm{c}}^{°}$ of H₂)]

              C+ O₂ $\to$CO₂,    $∆\mathrm{H}°$ = -94 kcal   ...(i)

           2H₂ + O₂ $\to$2H₂O,  $∆\mathrm{H}°$= -68 x 2 kcal         ...(ii)

     CH₄ + 2O₂ $\to$CO₂ + 2H₂O,  $∆\mathrm{H}°$=-213 kcal    .....(iii)

On adding eqs. (i) and(ii) and then subtracting eq (iii)

              = -[(-213)-(-94+2x-68)] kcal/mol

              = -[-213 + 230] = -17 kcal/mol

(a) q=$∆\mathrm{U} + \mathrm{W}$, its work done by the system.

    50=$∆$U + 30

 .^..   $∆$U = 20 cal

(d) In a cyclic process, $∆\mathrm{U}=0$

(d) The gaseous phase have more entropy and thus, $∆$S is +ve in (a) and (b). Also decrease in pressure increase disorder and thus, $∆$S is +ve in (c). In (d) the disorder decreases in liquid state due to the decrease in temperature. Thus, $∆$S= -ve

(b) If work done by the system is positive, then q=$∆$U+W. However, new terminology has revealed that work done by the system is negative and work done on the system is positive. Thus, according to this, q=$∆$E-W.