272708
A gas performs $0.320 \mathrm{~kJ}$ work on surrounding and absorbs $120 \mathrm{~J}$ of heat from the surrounding. Hence, change in internal energy is
1 $200 \mathrm{~J}$
2 $120.32 \mathrm{~J}$
3 $-200 \mathrm{~J}$
4 $440 \mathrm{~J}$
Explanation:
According to first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ Since, work is done on the surrounding So, $\quad \mathrm{W}=-0.320 \mathrm{~kJ}=-320 \mathrm{~J}$ $\mathrm{q}=120 \mathrm{~J}$ $\therefore \quad \Delta \mathrm{U}=120-320=-200 \mathrm{~J}$
Shift-I
Thermodynamics
272710
Calculate the work done during combustion of $0.138 \mathrm{~kg}$ of ethanol, $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})$ at $300 \mathrm{~K}$. Given: $\mathbf{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, molar mass of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$.
1 $-7482 \mathrm{~J}$
2 $7482 \mathrm{~J}$
3 $-2494 \mathrm{~J}$
4 $2494 \mathrm{~J}$
Explanation:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l)$ $\text { No. of moles of ethanol }=\frac{\text { Mass of ethanol }}{\text { Molar mass of ethanol }}$ $\qquad=\frac{138}{46}=3$ $3 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+9 \mathrm{O}_2 \longrightarrow 6 \mathrm{CO}_2+9 \mathrm{H}_2 \mathrm{O}$ $\Delta \mathrm{n}=6-9=-3$ $\text { Work }(\mathrm{W})=-\Delta \mathrm{nRT}$ $\mathrm{W}=-(-3) \times 8.314 \times 300$ $\mathrm{~W}=7482 \mathrm{~J}$
MHT CET-2018
Thermodynamics
272713
The work done during combustion of $9 \times 10^{-2}$ $\mathrm{kg}$ of ethane, $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})$ at $300 \mathrm{~K}$ is (Given $\mathrm{R}=$ $8.314 \mathrm{~J} \mathrm{deg}^{-1} \mathrm{~mol}^{-1}$, atomic mass $\mathrm{C}=12, \mathrm{H}=1$ )
1 $6.236 \mathrm{~kJ}$
2 $-6.236 \mathrm{~kJ}$
3 $18.71 \mathrm{~kJ}$
4 $-18.71 \mathrm{~kJ}$
Explanation:
$2 \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+7 \mathrm{O}_2(\mathrm{~g}) \longrightarrow$ $\Delta \mathrm{n}=2-(1+3.5)=-2.5$ $\mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(l)$ Molar mass of ethane $=2(12)+6(1)=30 \mathrm{~g} / \mathrm{mol}$ $9 \times 10^{-2} \mathrm{~kg}$ ethane $=9 \times 10^{-2} \mathrm{~kg} \times 1000 \mathrm{~g} / \mathrm{kg}=90 \mathrm{~g}$ $90 \mathrm{~g}$ ethane $=\frac{90 \mathrm{~g}}{30 \mathrm{~g} / \mathrm{mol}}=3 \mathrm{~mol}$ ethane For 1 mole of ethane, $\Delta \mathrm{n}=-2.5$ For 3 mole of ethane, $\Delta \mathrm{n}=-2.5 \times 3=-7.5$ Work done, W $=-\Delta \mathrm{nRT}$ $\mathrm{W}=-(-7.5) \times 8.314 \times 300$ $\mathrm{~W}=18707 \mathrm{~J}$ $\mathrm{~W}=18.71 \mathrm{~kJ}$
MHT CET-2017
Thermodynamics
272714
What is the amount of work done when 0.5 mole of methane, $\mathrm{CH}_4(\mathrm{~g})$, is subjected to combustion at $300 \mathrm{~K}$ ? (Given $R=8.314 \mathrm{JK}^{-1}$ $\mathrm{mol}^{-1}$ )
1 $-2494 \mathrm{~J}$
2 $-4988 \mathrm{~J}$
3 $+4988 \mathrm{~J}$
4 $+2494 \mathrm{~J}$
Explanation:
The balanced chemical equation for combustion reaction is, $\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ When 1 mole of methane undergoes combustion $\Delta \mathrm{n}=1-(1+2)=-2 \quad \therefore \Delta \mathrm{n}=$ Negative When 0.5 mole of methane undergo combustion $\Delta \mathrm{n}=-2 \times 0.5=-1$ At STP ( $1 \mathrm{~atm}$ and $273 \mathrm{~K}$ ) 1 mole of a gas occupies a volume of $22.4 \mathrm{~L}$. The Volume change $(\Delta \mathrm{V})=22.4 \mathrm{~L} \times \Delta \mathrm{n} \times \frac{300 \mathrm{~K}}{273 \mathrm{~K}}$ $\Delta \mathrm{V}=-24.62 \mathrm{~L}$ Work done $(\mathrm{W})=-\mathrm{P} \Delta \mathrm{V}$ $\mathrm{W}=-1 \mathrm{~atm} \times(-24.62 \mathrm{~L})$ $\mathrm{W}=+24.6 \mathrm{~L}$ atm $1 \mathrm{~L}$ atm $=101.3 \mathrm{~J}$ $\mathrm{W}=24.62 \mathrm{~atm} \times 101.3 \mathrm{~J}$ Hence, the amount of work done is $+2494 \mathrm{~J}$.
272708
A gas performs $0.320 \mathrm{~kJ}$ work on surrounding and absorbs $120 \mathrm{~J}$ of heat from the surrounding. Hence, change in internal energy is
1 $200 \mathrm{~J}$
2 $120.32 \mathrm{~J}$
3 $-200 \mathrm{~J}$
4 $440 \mathrm{~J}$
Explanation:
According to first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ Since, work is done on the surrounding So, $\quad \mathrm{W}=-0.320 \mathrm{~kJ}=-320 \mathrm{~J}$ $\mathrm{q}=120 \mathrm{~J}$ $\therefore \quad \Delta \mathrm{U}=120-320=-200 \mathrm{~J}$
Shift-I
Thermodynamics
272710
Calculate the work done during combustion of $0.138 \mathrm{~kg}$ of ethanol, $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})$ at $300 \mathrm{~K}$. Given: $\mathbf{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, molar mass of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$.
1 $-7482 \mathrm{~J}$
2 $7482 \mathrm{~J}$
3 $-2494 \mathrm{~J}$
4 $2494 \mathrm{~J}$
Explanation:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l)$ $\text { No. of moles of ethanol }=\frac{\text { Mass of ethanol }}{\text { Molar mass of ethanol }}$ $\qquad=\frac{138}{46}=3$ $3 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+9 \mathrm{O}_2 \longrightarrow 6 \mathrm{CO}_2+9 \mathrm{H}_2 \mathrm{O}$ $\Delta \mathrm{n}=6-9=-3$ $\text { Work }(\mathrm{W})=-\Delta \mathrm{nRT}$ $\mathrm{W}=-(-3) \times 8.314 \times 300$ $\mathrm{~W}=7482 \mathrm{~J}$
MHT CET-2018
Thermodynamics
272713
The work done during combustion of $9 \times 10^{-2}$ $\mathrm{kg}$ of ethane, $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})$ at $300 \mathrm{~K}$ is (Given $\mathrm{R}=$ $8.314 \mathrm{~J} \mathrm{deg}^{-1} \mathrm{~mol}^{-1}$, atomic mass $\mathrm{C}=12, \mathrm{H}=1$ )
1 $6.236 \mathrm{~kJ}$
2 $-6.236 \mathrm{~kJ}$
3 $18.71 \mathrm{~kJ}$
4 $-18.71 \mathrm{~kJ}$
Explanation:
$2 \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+7 \mathrm{O}_2(\mathrm{~g}) \longrightarrow$ $\Delta \mathrm{n}=2-(1+3.5)=-2.5$ $\mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(l)$ Molar mass of ethane $=2(12)+6(1)=30 \mathrm{~g} / \mathrm{mol}$ $9 \times 10^{-2} \mathrm{~kg}$ ethane $=9 \times 10^{-2} \mathrm{~kg} \times 1000 \mathrm{~g} / \mathrm{kg}=90 \mathrm{~g}$ $90 \mathrm{~g}$ ethane $=\frac{90 \mathrm{~g}}{30 \mathrm{~g} / \mathrm{mol}}=3 \mathrm{~mol}$ ethane For 1 mole of ethane, $\Delta \mathrm{n}=-2.5$ For 3 mole of ethane, $\Delta \mathrm{n}=-2.5 \times 3=-7.5$ Work done, W $=-\Delta \mathrm{nRT}$ $\mathrm{W}=-(-7.5) \times 8.314 \times 300$ $\mathrm{~W}=18707 \mathrm{~J}$ $\mathrm{~W}=18.71 \mathrm{~kJ}$
MHT CET-2017
Thermodynamics
272714
What is the amount of work done when 0.5 mole of methane, $\mathrm{CH}_4(\mathrm{~g})$, is subjected to combustion at $300 \mathrm{~K}$ ? (Given $R=8.314 \mathrm{JK}^{-1}$ $\mathrm{mol}^{-1}$ )
1 $-2494 \mathrm{~J}$
2 $-4988 \mathrm{~J}$
3 $+4988 \mathrm{~J}$
4 $+2494 \mathrm{~J}$
Explanation:
The balanced chemical equation for combustion reaction is, $\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ When 1 mole of methane undergoes combustion $\Delta \mathrm{n}=1-(1+2)=-2 \quad \therefore \Delta \mathrm{n}=$ Negative When 0.5 mole of methane undergo combustion $\Delta \mathrm{n}=-2 \times 0.5=-1$ At STP ( $1 \mathrm{~atm}$ and $273 \mathrm{~K}$ ) 1 mole of a gas occupies a volume of $22.4 \mathrm{~L}$. The Volume change $(\Delta \mathrm{V})=22.4 \mathrm{~L} \times \Delta \mathrm{n} \times \frac{300 \mathrm{~K}}{273 \mathrm{~K}}$ $\Delta \mathrm{V}=-24.62 \mathrm{~L}$ Work done $(\mathrm{W})=-\mathrm{P} \Delta \mathrm{V}$ $\mathrm{W}=-1 \mathrm{~atm} \times(-24.62 \mathrm{~L})$ $\mathrm{W}=+24.6 \mathrm{~L}$ atm $1 \mathrm{~L}$ atm $=101.3 \mathrm{~J}$ $\mathrm{W}=24.62 \mathrm{~atm} \times 101.3 \mathrm{~J}$ Hence, the amount of work done is $+2494 \mathrm{~J}$.
272708
A gas performs $0.320 \mathrm{~kJ}$ work on surrounding and absorbs $120 \mathrm{~J}$ of heat from the surrounding. Hence, change in internal energy is
1 $200 \mathrm{~J}$
2 $120.32 \mathrm{~J}$
3 $-200 \mathrm{~J}$
4 $440 \mathrm{~J}$
Explanation:
According to first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ Since, work is done on the surrounding So, $\quad \mathrm{W}=-0.320 \mathrm{~kJ}=-320 \mathrm{~J}$ $\mathrm{q}=120 \mathrm{~J}$ $\therefore \quad \Delta \mathrm{U}=120-320=-200 \mathrm{~J}$
Shift-I
Thermodynamics
272710
Calculate the work done during combustion of $0.138 \mathrm{~kg}$ of ethanol, $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})$ at $300 \mathrm{~K}$. Given: $\mathbf{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, molar mass of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$.
1 $-7482 \mathrm{~J}$
2 $7482 \mathrm{~J}$
3 $-2494 \mathrm{~J}$
4 $2494 \mathrm{~J}$
Explanation:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l)$ $\text { No. of moles of ethanol }=\frac{\text { Mass of ethanol }}{\text { Molar mass of ethanol }}$ $\qquad=\frac{138}{46}=3$ $3 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+9 \mathrm{O}_2 \longrightarrow 6 \mathrm{CO}_2+9 \mathrm{H}_2 \mathrm{O}$ $\Delta \mathrm{n}=6-9=-3$ $\text { Work }(\mathrm{W})=-\Delta \mathrm{nRT}$ $\mathrm{W}=-(-3) \times 8.314 \times 300$ $\mathrm{~W}=7482 \mathrm{~J}$
MHT CET-2018
Thermodynamics
272713
The work done during combustion of $9 \times 10^{-2}$ $\mathrm{kg}$ of ethane, $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})$ at $300 \mathrm{~K}$ is (Given $\mathrm{R}=$ $8.314 \mathrm{~J} \mathrm{deg}^{-1} \mathrm{~mol}^{-1}$, atomic mass $\mathrm{C}=12, \mathrm{H}=1$ )
1 $6.236 \mathrm{~kJ}$
2 $-6.236 \mathrm{~kJ}$
3 $18.71 \mathrm{~kJ}$
4 $-18.71 \mathrm{~kJ}$
Explanation:
$2 \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+7 \mathrm{O}_2(\mathrm{~g}) \longrightarrow$ $\Delta \mathrm{n}=2-(1+3.5)=-2.5$ $\mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(l)$ Molar mass of ethane $=2(12)+6(1)=30 \mathrm{~g} / \mathrm{mol}$ $9 \times 10^{-2} \mathrm{~kg}$ ethane $=9 \times 10^{-2} \mathrm{~kg} \times 1000 \mathrm{~g} / \mathrm{kg}=90 \mathrm{~g}$ $90 \mathrm{~g}$ ethane $=\frac{90 \mathrm{~g}}{30 \mathrm{~g} / \mathrm{mol}}=3 \mathrm{~mol}$ ethane For 1 mole of ethane, $\Delta \mathrm{n}=-2.5$ For 3 mole of ethane, $\Delta \mathrm{n}=-2.5 \times 3=-7.5$ Work done, W $=-\Delta \mathrm{nRT}$ $\mathrm{W}=-(-7.5) \times 8.314 \times 300$ $\mathrm{~W}=18707 \mathrm{~J}$ $\mathrm{~W}=18.71 \mathrm{~kJ}$
MHT CET-2017
Thermodynamics
272714
What is the amount of work done when 0.5 mole of methane, $\mathrm{CH}_4(\mathrm{~g})$, is subjected to combustion at $300 \mathrm{~K}$ ? (Given $R=8.314 \mathrm{JK}^{-1}$ $\mathrm{mol}^{-1}$ )
1 $-2494 \mathrm{~J}$
2 $-4988 \mathrm{~J}$
3 $+4988 \mathrm{~J}$
4 $+2494 \mathrm{~J}$
Explanation:
The balanced chemical equation for combustion reaction is, $\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ When 1 mole of methane undergoes combustion $\Delta \mathrm{n}=1-(1+2)=-2 \quad \therefore \Delta \mathrm{n}=$ Negative When 0.5 mole of methane undergo combustion $\Delta \mathrm{n}=-2 \times 0.5=-1$ At STP ( $1 \mathrm{~atm}$ and $273 \mathrm{~K}$ ) 1 mole of a gas occupies a volume of $22.4 \mathrm{~L}$. The Volume change $(\Delta \mathrm{V})=22.4 \mathrm{~L} \times \Delta \mathrm{n} \times \frac{300 \mathrm{~K}}{273 \mathrm{~K}}$ $\Delta \mathrm{V}=-24.62 \mathrm{~L}$ Work done $(\mathrm{W})=-\mathrm{P} \Delta \mathrm{V}$ $\mathrm{W}=-1 \mathrm{~atm} \times(-24.62 \mathrm{~L})$ $\mathrm{W}=+24.6 \mathrm{~L}$ atm $1 \mathrm{~L}$ atm $=101.3 \mathrm{~J}$ $\mathrm{W}=24.62 \mathrm{~atm} \times 101.3 \mathrm{~J}$ Hence, the amount of work done is $+2494 \mathrm{~J}$.
272708
A gas performs $0.320 \mathrm{~kJ}$ work on surrounding and absorbs $120 \mathrm{~J}$ of heat from the surrounding. Hence, change in internal energy is
1 $200 \mathrm{~J}$
2 $120.32 \mathrm{~J}$
3 $-200 \mathrm{~J}$
4 $440 \mathrm{~J}$
Explanation:
According to first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ Since, work is done on the surrounding So, $\quad \mathrm{W}=-0.320 \mathrm{~kJ}=-320 \mathrm{~J}$ $\mathrm{q}=120 \mathrm{~J}$ $\therefore \quad \Delta \mathrm{U}=120-320=-200 \mathrm{~J}$
Shift-I
Thermodynamics
272710
Calculate the work done during combustion of $0.138 \mathrm{~kg}$ of ethanol, $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})$ at $300 \mathrm{~K}$. Given: $\mathbf{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, molar mass of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$.
1 $-7482 \mathrm{~J}$
2 $7482 \mathrm{~J}$
3 $-2494 \mathrm{~J}$
4 $2494 \mathrm{~J}$
Explanation:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l)$ $\text { No. of moles of ethanol }=\frac{\text { Mass of ethanol }}{\text { Molar mass of ethanol }}$ $\qquad=\frac{138}{46}=3$ $3 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+9 \mathrm{O}_2 \longrightarrow 6 \mathrm{CO}_2+9 \mathrm{H}_2 \mathrm{O}$ $\Delta \mathrm{n}=6-9=-3$ $\text { Work }(\mathrm{W})=-\Delta \mathrm{nRT}$ $\mathrm{W}=-(-3) \times 8.314 \times 300$ $\mathrm{~W}=7482 \mathrm{~J}$
MHT CET-2018
Thermodynamics
272713
The work done during combustion of $9 \times 10^{-2}$ $\mathrm{kg}$ of ethane, $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})$ at $300 \mathrm{~K}$ is (Given $\mathrm{R}=$ $8.314 \mathrm{~J} \mathrm{deg}^{-1} \mathrm{~mol}^{-1}$, atomic mass $\mathrm{C}=12, \mathrm{H}=1$ )
1 $6.236 \mathrm{~kJ}$
2 $-6.236 \mathrm{~kJ}$
3 $18.71 \mathrm{~kJ}$
4 $-18.71 \mathrm{~kJ}$
Explanation:
$2 \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+7 \mathrm{O}_2(\mathrm{~g}) \longrightarrow$ $\Delta \mathrm{n}=2-(1+3.5)=-2.5$ $\mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(l)$ Molar mass of ethane $=2(12)+6(1)=30 \mathrm{~g} / \mathrm{mol}$ $9 \times 10^{-2} \mathrm{~kg}$ ethane $=9 \times 10^{-2} \mathrm{~kg} \times 1000 \mathrm{~g} / \mathrm{kg}=90 \mathrm{~g}$ $90 \mathrm{~g}$ ethane $=\frac{90 \mathrm{~g}}{30 \mathrm{~g} / \mathrm{mol}}=3 \mathrm{~mol}$ ethane For 1 mole of ethane, $\Delta \mathrm{n}=-2.5$ For 3 mole of ethane, $\Delta \mathrm{n}=-2.5 \times 3=-7.5$ Work done, W $=-\Delta \mathrm{nRT}$ $\mathrm{W}=-(-7.5) \times 8.314 \times 300$ $\mathrm{~W}=18707 \mathrm{~J}$ $\mathrm{~W}=18.71 \mathrm{~kJ}$
MHT CET-2017
Thermodynamics
272714
What is the amount of work done when 0.5 mole of methane, $\mathrm{CH}_4(\mathrm{~g})$, is subjected to combustion at $300 \mathrm{~K}$ ? (Given $R=8.314 \mathrm{JK}^{-1}$ $\mathrm{mol}^{-1}$ )
1 $-2494 \mathrm{~J}$
2 $-4988 \mathrm{~J}$
3 $+4988 \mathrm{~J}$
4 $+2494 \mathrm{~J}$
Explanation:
The balanced chemical equation for combustion reaction is, $\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ When 1 mole of methane undergoes combustion $\Delta \mathrm{n}=1-(1+2)=-2 \quad \therefore \Delta \mathrm{n}=$ Negative When 0.5 mole of methane undergo combustion $\Delta \mathrm{n}=-2 \times 0.5=-1$ At STP ( $1 \mathrm{~atm}$ and $273 \mathrm{~K}$ ) 1 mole of a gas occupies a volume of $22.4 \mathrm{~L}$. The Volume change $(\Delta \mathrm{V})=22.4 \mathrm{~L} \times \Delta \mathrm{n} \times \frac{300 \mathrm{~K}}{273 \mathrm{~K}}$ $\Delta \mathrm{V}=-24.62 \mathrm{~L}$ Work done $(\mathrm{W})=-\mathrm{P} \Delta \mathrm{V}$ $\mathrm{W}=-1 \mathrm{~atm} \times(-24.62 \mathrm{~L})$ $\mathrm{W}=+24.6 \mathrm{~L}$ atm $1 \mathrm{~L}$ atm $=101.3 \mathrm{~J}$ $\mathrm{W}=24.62 \mathrm{~atm} \times 101.3 \mathrm{~J}$ Hence, the amount of work done is $+2494 \mathrm{~J}$.
