NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
272715
Given: $\mathbf{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, the work done during combustion of $0.090 \mathrm{~kg}$ of ethane (molar mass $=30$ ) at $300 \mathrm{~K}$ is
1 $-18.7 \mathrm{~kJ}$
2 $18.7 \mathrm{~kJ}$
3 $6.234 \mathrm{~kJ}$
4 $-6.234 \mathrm{~kJ}$
Explanation:
The combustion of $\mathrm{C}_2 \mathrm{H}_6$ is as, $\mathrm{C}_2 \mathrm{H}_6+7 / 2 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}$ No of moles of $\mathrm{C}_2 \mathrm{H}_6=\frac{\text { Mass }}{\text { Molar mass }}$ $=(0.090 \mathrm{~kg})\left(.30 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}{ }^{-1}\right)$ Change in number of moles of gas is, $\Delta \mathrm{n}=\Sigma \mathrm{n}_{\mathrm{g} \text { (produced) }}-\Sigma \mathrm{n}_{\mathrm{g} \text { (reactaut) }}$ $\Delta \mathrm{n}=2-(1+7 / 2)=-2.5 \quad \therefore \Delta \mathrm{n}=\text { Negative }$ Now, work done $(\mathrm{W})=-\mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{nRT}$ $W=-(-2.5 \times 8.314 \times 300)$ $=6235.5 \mathrm{~J}$ For one mole combustion, $\mathrm{W}=6235.5 \mathrm{~J}$ For 3 moles, $\mathrm{W} =3 \times 6235.5 \mathrm{~J}$ $=18706.5 \mathrm{~J}=18.7 \mathrm{~kJ}$
MHT CET-2015
Thermodynamics
272716
The work done when two mole of an ideal gas is compressed form a volume of $5 \mathrm{~m}^3$ to $1 \mathrm{dm}^3$ at $300 \mathrm{~K}$, under a pressure of $100 \mathrm{kPa}$ is
272718
At the same conditions of pressure, volume and temperature, work done is maximum for which gas if all gases have equal masses?
1 $\mathrm{NH}_3$
2 $\mathrm{N}_2$
3 $\mathrm{Cl}_2$
4 $\mathrm{H}_2 \mathrm{~S}$]#
Explanation:
When $P, V$ and $T$ are same and mass is also small work done depends only on molecular mass. $\mathrm{W} \propto \frac{1}{\mathrm{M}}$ (Where, $M=$ molecular mass) $\mathrm{NH}_3 \rightarrow 14+3=17 \mathrm{~g} /$ mole $\mathrm{N}_2 \rightarrow 14 \times 2=28 \mathrm{~g} /$ mole $\mathrm{Cl}_2 \rightarrow 35.5 \times 2=71 \mathrm{~g} / \mathrm{mole}$ $\mathrm{H}_2 \mathrm{~S} \rightarrow 2+32=34 \mathrm{~g} / \mathrm{mole}$ Among the given gases, $\mathrm{NH}_3$ has lowest molecular mass. So work done is maximum.
272715
Given: $\mathbf{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, the work done during combustion of $0.090 \mathrm{~kg}$ of ethane (molar mass $=30$ ) at $300 \mathrm{~K}$ is
1 $-18.7 \mathrm{~kJ}$
2 $18.7 \mathrm{~kJ}$
3 $6.234 \mathrm{~kJ}$
4 $-6.234 \mathrm{~kJ}$
Explanation:
The combustion of $\mathrm{C}_2 \mathrm{H}_6$ is as, $\mathrm{C}_2 \mathrm{H}_6+7 / 2 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}$ No of moles of $\mathrm{C}_2 \mathrm{H}_6=\frac{\text { Mass }}{\text { Molar mass }}$ $=(0.090 \mathrm{~kg})\left(.30 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}{ }^{-1}\right)$ Change in number of moles of gas is, $\Delta \mathrm{n}=\Sigma \mathrm{n}_{\mathrm{g} \text { (produced) }}-\Sigma \mathrm{n}_{\mathrm{g} \text { (reactaut) }}$ $\Delta \mathrm{n}=2-(1+7 / 2)=-2.5 \quad \therefore \Delta \mathrm{n}=\text { Negative }$ Now, work done $(\mathrm{W})=-\mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{nRT}$ $W=-(-2.5 \times 8.314 \times 300)$ $=6235.5 \mathrm{~J}$ For one mole combustion, $\mathrm{W}=6235.5 \mathrm{~J}$ For 3 moles, $\mathrm{W} =3 \times 6235.5 \mathrm{~J}$ $=18706.5 \mathrm{~J}=18.7 \mathrm{~kJ}$
MHT CET-2015
Thermodynamics
272716
The work done when two mole of an ideal gas is compressed form a volume of $5 \mathrm{~m}^3$ to $1 \mathrm{dm}^3$ at $300 \mathrm{~K}$, under a pressure of $100 \mathrm{kPa}$ is
272718
At the same conditions of pressure, volume and temperature, work done is maximum for which gas if all gases have equal masses?
1 $\mathrm{NH}_3$
2 $\mathrm{N}_2$
3 $\mathrm{Cl}_2$
4 $\mathrm{H}_2 \mathrm{~S}$]#
Explanation:
When $P, V$ and $T$ are same and mass is also small work done depends only on molecular mass. $\mathrm{W} \propto \frac{1}{\mathrm{M}}$ (Where, $M=$ molecular mass) $\mathrm{NH}_3 \rightarrow 14+3=17 \mathrm{~g} /$ mole $\mathrm{N}_2 \rightarrow 14 \times 2=28 \mathrm{~g} /$ mole $\mathrm{Cl}_2 \rightarrow 35.5 \times 2=71 \mathrm{~g} / \mathrm{mole}$ $\mathrm{H}_2 \mathrm{~S} \rightarrow 2+32=34 \mathrm{~g} / \mathrm{mole}$ Among the given gases, $\mathrm{NH}_3$ has lowest molecular mass. So work done is maximum.
272715
Given: $\mathbf{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, the work done during combustion of $0.090 \mathrm{~kg}$ of ethane (molar mass $=30$ ) at $300 \mathrm{~K}$ is
1 $-18.7 \mathrm{~kJ}$
2 $18.7 \mathrm{~kJ}$
3 $6.234 \mathrm{~kJ}$
4 $-6.234 \mathrm{~kJ}$
Explanation:
The combustion of $\mathrm{C}_2 \mathrm{H}_6$ is as, $\mathrm{C}_2 \mathrm{H}_6+7 / 2 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}$ No of moles of $\mathrm{C}_2 \mathrm{H}_6=\frac{\text { Mass }}{\text { Molar mass }}$ $=(0.090 \mathrm{~kg})\left(.30 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}{ }^{-1}\right)$ Change in number of moles of gas is, $\Delta \mathrm{n}=\Sigma \mathrm{n}_{\mathrm{g} \text { (produced) }}-\Sigma \mathrm{n}_{\mathrm{g} \text { (reactaut) }}$ $\Delta \mathrm{n}=2-(1+7 / 2)=-2.5 \quad \therefore \Delta \mathrm{n}=\text { Negative }$ Now, work done $(\mathrm{W})=-\mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{nRT}$ $W=-(-2.5 \times 8.314 \times 300)$ $=6235.5 \mathrm{~J}$ For one mole combustion, $\mathrm{W}=6235.5 \mathrm{~J}$ For 3 moles, $\mathrm{W} =3 \times 6235.5 \mathrm{~J}$ $=18706.5 \mathrm{~J}=18.7 \mathrm{~kJ}$
MHT CET-2015
Thermodynamics
272716
The work done when two mole of an ideal gas is compressed form a volume of $5 \mathrm{~m}^3$ to $1 \mathrm{dm}^3$ at $300 \mathrm{~K}$, under a pressure of $100 \mathrm{kPa}$ is
272718
At the same conditions of pressure, volume and temperature, work done is maximum for which gas if all gases have equal masses?
1 $\mathrm{NH}_3$
2 $\mathrm{N}_2$
3 $\mathrm{Cl}_2$
4 $\mathrm{H}_2 \mathrm{~S}$]#
Explanation:
When $P, V$ and $T$ are same and mass is also small work done depends only on molecular mass. $\mathrm{W} \propto \frac{1}{\mathrm{M}}$ (Where, $M=$ molecular mass) $\mathrm{NH}_3 \rightarrow 14+3=17 \mathrm{~g} /$ mole $\mathrm{N}_2 \rightarrow 14 \times 2=28 \mathrm{~g} /$ mole $\mathrm{Cl}_2 \rightarrow 35.5 \times 2=71 \mathrm{~g} / \mathrm{mole}$ $\mathrm{H}_2 \mathrm{~S} \rightarrow 2+32=34 \mathrm{~g} / \mathrm{mole}$ Among the given gases, $\mathrm{NH}_3$ has lowest molecular mass. So work done is maximum.
272715
Given: $\mathbf{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, the work done during combustion of $0.090 \mathrm{~kg}$ of ethane (molar mass $=30$ ) at $300 \mathrm{~K}$ is
1 $-18.7 \mathrm{~kJ}$
2 $18.7 \mathrm{~kJ}$
3 $6.234 \mathrm{~kJ}$
4 $-6.234 \mathrm{~kJ}$
Explanation:
The combustion of $\mathrm{C}_2 \mathrm{H}_6$ is as, $\mathrm{C}_2 \mathrm{H}_6+7 / 2 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}$ No of moles of $\mathrm{C}_2 \mathrm{H}_6=\frac{\text { Mass }}{\text { Molar mass }}$ $=(0.090 \mathrm{~kg})\left(.30 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}{ }^{-1}\right)$ Change in number of moles of gas is, $\Delta \mathrm{n}=\Sigma \mathrm{n}_{\mathrm{g} \text { (produced) }}-\Sigma \mathrm{n}_{\mathrm{g} \text { (reactaut) }}$ $\Delta \mathrm{n}=2-(1+7 / 2)=-2.5 \quad \therefore \Delta \mathrm{n}=\text { Negative }$ Now, work done $(\mathrm{W})=-\mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{nRT}$ $W=-(-2.5 \times 8.314 \times 300)$ $=6235.5 \mathrm{~J}$ For one mole combustion, $\mathrm{W}=6235.5 \mathrm{~J}$ For 3 moles, $\mathrm{W} =3 \times 6235.5 \mathrm{~J}$ $=18706.5 \mathrm{~J}=18.7 \mathrm{~kJ}$
MHT CET-2015
Thermodynamics
272716
The work done when two mole of an ideal gas is compressed form a volume of $5 \mathrm{~m}^3$ to $1 \mathrm{dm}^3$ at $300 \mathrm{~K}$, under a pressure of $100 \mathrm{kPa}$ is
272718
At the same conditions of pressure, volume and temperature, work done is maximum for which gas if all gases have equal masses?
1 $\mathrm{NH}_3$
2 $\mathrm{N}_2$
3 $\mathrm{Cl}_2$
4 $\mathrm{H}_2 \mathrm{~S}$]#
Explanation:
When $P, V$ and $T$ are same and mass is also small work done depends only on molecular mass. $\mathrm{W} \propto \frac{1}{\mathrm{M}}$ (Where, $M=$ molecular mass) $\mathrm{NH}_3 \rightarrow 14+3=17 \mathrm{~g} /$ mole $\mathrm{N}_2 \rightarrow 14 \times 2=28 \mathrm{~g} /$ mole $\mathrm{Cl}_2 \rightarrow 35.5 \times 2=71 \mathrm{~g} / \mathrm{mole}$ $\mathrm{H}_2 \mathrm{~S} \rightarrow 2+32=34 \mathrm{~g} / \mathrm{mole}$ Among the given gases, $\mathrm{NH}_3$ has lowest molecular mass. So work done is maximum.
