MCQ
For an isothermal reversible expansion process, the value of $q$ can be calculated by the expression
  • $q = 2.303\,nRT\,\log \,\frac{{{V_2}}}{{{V_1}}}$
  • B
    $q = -2.303\,nRT\,\log \,\frac{{{V_2}}}{{{V_1}}}$
  • C
    $q =  - 2.303\,nR\,\log \,\frac{{{V_1}}}{{{V_2}}}$
  • D
    $q =  - {P_{\exp }}\,nRT\,\log \,\frac{{{V_1}}}{{{V_2}}}$

Answer

Correct option: A.
$q = 2.303\,nRT\,\log \,\frac{{{V_2}}}{{{V_1}}}$
a
$\Delta \mathrm{E}=\mathrm{q}+\mathrm{w}$

$0=\mathrm{q}+\mathrm{w}$

$\mathrm{q}=-\mathrm{w}-\left(-2.303\, \mathrm{n} \mathrm{R} \mathrm{T} \,\log\, \frac{\mathrm{V}_{2}}{\mathrm{V}_{1}}\right)$

$=2.303\, \mathrm{nRT} \,\log \,\frac{\mathrm{V}_{2}}{\mathrm{V}_{1}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free