The temperature of 4 moles of an ideal gas is raised from $300 \mathrm{~K}$ to $350 \mathrm{~K}$. What is…

The temperature of 4 moles of an ideal gas is raised from $300 \mathrm{~K}$ to $350 \mathrm{~K}$. What is the vahe of $\Delta H-\Delta E$ for this process ? $\quad\left(R=8.3 \mathrm{~J} \mathrm{~mol}^{-1} K^{-1}ight)$
  1. 0
  2. $415 \mathrm{~J}$
  3. $41.5 \mathrm{~J}$
  4. $\quad 1660 \mathrm{~J}$

Solution

$\mathbf{H}=\mathrm{E}+\mathbf{P V}$
$\mathbf{H}=\mathbf{E}+\mathbf{n R T}$
$\mathrm{H}_{2}=\mathrm{E}_{2}+\mathrm{nRT}_{2}$
$\mathbf{H}_{1}=\mathbf{E}_{1}+\mathbf{n} \mathbf{R} \mathbf{T}_{1}$
$\therefore\left(\mathbf{H}_{2}-\mathbf{H}_{1}ight)=\left(\mathbf{E}_{2}-\mathbf{E}_{1}ight)+\mathbf{n R}\left(\mathbf{T}_{2}-\mathbf{T}_{1}ight)$
$\Delta \mathrm{H}-\Delta \mathrm{E}=\mathbf{n R} \mathrm{AT}=4 \times 8.3 \times 50=1660 \mathrm{~J}$
Thus, (d) *

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more THERMODYNAMICS questions on Aicharya