JEE Advance - Chemistry (2016 - Paper 1 Offline - No. 4)
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to
2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surrounding ($$\Delta$$Ssurr)in
JK–1 is (1L atm = 101.3 J)
5.763
1.013
– 1.013
– 5.763
Explanation
Work done in an irreversible isothermal process is given by :
$$ \begin{aligned} \mathrm{W}_{\mathrm{irr}} & =-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & =-3.0 \mathrm{~atm}(2.0 \mathrm{~L}-1.0 \mathrm{~L}) \\\\ & =-3.0 \times 1.0 \mathrm{~atm} \mathrm{~L} \\\\ \mathrm{~W}_{\mathrm{irr}} & =-3.0 \mathrm{~L} \mathrm{~atm} \\\\ \mathrm{~W}_{\mathrm{irr}} & =-3.0 \times 101.3 \mathrm{~J} \,\,\,\,\,\,\,\,\,\ [1 \mathrm{~L} \mathrm{~atm}=101.3] \\\\ & =-303.9 \mathrm{~J} \end{aligned} $$
At a constant pressure,
$$ \begin{aligned} & \Delta \mathrm{U}=\mathrm{q}_{\mathrm{p}}+\mathrm{W}_{\text {irr }}=\mathrm{q}_{\mathrm{p}}-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & \Delta \mathrm{U}=\Delta \mathrm{H}-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & \Delta \mathrm{H}=\Delta \mathrm{U}+\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \end{aligned} $$
Since, reaction takes place at constant temperature; hence, $\Delta \mathrm{U}=0$.
$$ \begin{aligned} \Delta \mathrm{H} & =\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right)=-\mathrm{W}_{\mathrm{irr}}=303.9 \mathrm{~J} \\\\ \Delta \mathrm{H}_{\mathrm{sys}} & =303.9 \mathrm{~J} \\\\ \Delta \mathrm{S}_{\text {sys }} & =\frac{\Delta \mathrm{H}_{\text {sys }}}{\mathrm{T}} \text { and } \\\\ \end{aligned} $$
Finally, the change in entropy of the surroundings (ΔSsurr) is equal to the negative of the change in entropy of the system, because the system and its surroundings form an isolated system :
ΔSsurr = -ΔSsys
$$ \begin{aligned} \Delta \mathrm{S}_{\text {surr }} & =-\frac{\Delta \mathrm{H}_{\text {sys }}}{\mathrm{T}}=-\frac{303.9 \mathrm{~J}}{300 \mathrm{~K}} \\\\ & =-1.013 \mathrm{~J} \mathrm{~K}^{-1} \end{aligned} $$
$$ \begin{aligned} \mathrm{W}_{\mathrm{irr}} & =-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & =-3.0 \mathrm{~atm}(2.0 \mathrm{~L}-1.0 \mathrm{~L}) \\\\ & =-3.0 \times 1.0 \mathrm{~atm} \mathrm{~L} \\\\ \mathrm{~W}_{\mathrm{irr}} & =-3.0 \mathrm{~L} \mathrm{~atm} \\\\ \mathrm{~W}_{\mathrm{irr}} & =-3.0 \times 101.3 \mathrm{~J} \,\,\,\,\,\,\,\,\,\ [1 \mathrm{~L} \mathrm{~atm}=101.3] \\\\ & =-303.9 \mathrm{~J} \end{aligned} $$
At a constant pressure,
$$ \begin{aligned} & \Delta \mathrm{U}=\mathrm{q}_{\mathrm{p}}+\mathrm{W}_{\text {irr }}=\mathrm{q}_{\mathrm{p}}-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & \Delta \mathrm{U}=\Delta \mathrm{H}-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & \Delta \mathrm{H}=\Delta \mathrm{U}+\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \end{aligned} $$
Since, reaction takes place at constant temperature; hence, $\Delta \mathrm{U}=0$.
$$ \begin{aligned} \Delta \mathrm{H} & =\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right)=-\mathrm{W}_{\mathrm{irr}}=303.9 \mathrm{~J} \\\\ \Delta \mathrm{H}_{\mathrm{sys}} & =303.9 \mathrm{~J} \\\\ \Delta \mathrm{S}_{\text {sys }} & =\frac{\Delta \mathrm{H}_{\text {sys }}}{\mathrm{T}} \text { and } \\\\ \end{aligned} $$
Finally, the change in entropy of the surroundings (ΔSsurr) is equal to the negative of the change in entropy of the system, because the system and its surroundings form an isolated system :
ΔSsurr = -ΔSsys
$$ \begin{aligned} \Delta \mathrm{S}_{\text {surr }} & =-\frac{\Delta \mathrm{H}_{\text {sys }}}{\mathrm{T}}=-\frac{303.9 \mathrm{~J}}{300 \mathrm{~K}} \\\\ & =-1.013 \mathrm{~J} \mathrm{~K}^{-1} \end{aligned} $$
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