272683
Temperature of 5 moles of a gas is decreased by $2 \mathrm{~K}$ at constant pressure. Indicate the correct statement
1 Work done by gas is $=5 \mathrm{R}$
2 Work done over the gas is $=10 \mathrm{R}$
3 Work done by the gas $=10 \mathrm{R}$
4 Work done $=0$
Explanation:
For 5 moles of gas at temperature T, $\mathrm{PV}_1=5 \mathrm{RT}$ For 5 moles of gas at temperature $(T-2)$, $\mathrm{PV}_2=5 \mathrm{R}(\mathrm{T}-2) \ldots \text { (ii) }$ $\therefore \mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)=5 \mathrm{R}(\mathrm{T}-2-\mathrm{T}) ;$ $\mathrm{P} \Delta \mathrm{V}=-10 \mathrm{R},-\mathrm{P} \Delta \mathrm{V}=10 \mathrm{R}$ When $\Delta \mathrm{V}$ is negative, $\mathrm{W}$ is +ve.
BITSAT-2007
Thermodynamics
272672
Two moles of an ideal gas are compressed isothermally $\left(100^{\circ} \mathrm{C}\right)$ and reversibly from a pressure of $10 \mathrm{~atm}$ to $25 \mathrm{~atm}$. Then the free energy change is:
1 $+15.482 \mathrm{~kJ}$
2 $+10.462 \mathrm{~kJ}$
3 $+5.684 \mathrm{~kJ}$
4 $+3.3642 \mathrm{~kJ}$
Explanation:
For isothermal reversible process: $\mathrm{W}_{\text {rev }} =-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{P}_1}{P_2}\right)=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_1}{\mathrm{~V}_2}$ $=-2.303 \times 2 \times 8.314 \times 373 \times \log \left(\frac{10}{25}\right)=+5684.08 \mathrm{~J}$ $\mathrm{W}_{\text {rev }}=+5.684 \mathrm{~kJ}$ Since, it is a measure of free energy $\therefore$ & $\Delta \mathrm{G}=\mathrm{W}_{\text {rev }}=\mathrm{W}_{\text {max }}$ $\therefore$ & $\Delta \mathrm{G}=+5.684 \mathrm{~kJ}$
AIIMS-2001
Thermodynamics
272673
A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 $\mathrm{dm}^3$ to a volume of $20 \mathrm{dm}^3$. It absorbs $800 \mathrm{~J}$ of thermal energy from its surrounding. The $\Delta \mathrm{U}$ is:
1 $-312 \mathrm{~J}$
2 $+123 \mathrm{~J}$
3 $-213 \mathrm{~J}$
4 $+231 \mathrm{~J}$
Explanation:
In general $\mathrm{W}=-\int_{\mathrm{V}_1}^{\mathrm{V}_2} \mathrm{PdV}=-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)$ $\mathrm{W}=-(1 \mathrm{~atm})(20-10)=-10 \mathrm{dm}^3 \mathrm{~atm}$ $\qquad\left[\begin{array}{l} \because 1 \mathrm{dm}^3=1 \mathrm{~L} \\ 1 \mathrm{atmL}=101.325 \mathrm{~J} \end{array}\right]$ $=-10 \times 101.325=-1013.25 \mathrm{~J} \approx-1013 \mathrm{~J}$ $\text { According to } 1^{\text {st }} \text { law of thermodynamics, }$ $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$ According to $1^{\text {st }}$ law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$
BCECE-2007
Thermodynamics
272675
When the temperature of 2 moles of an ideal gas is increased by $20^{\circ} \mathrm{C}$ at constant pressure. Find the work involved in the process.
272676
The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at $25^{\circ} \mathrm{C}$ is
272683
Temperature of 5 moles of a gas is decreased by $2 \mathrm{~K}$ at constant pressure. Indicate the correct statement
1 Work done by gas is $=5 \mathrm{R}$
2 Work done over the gas is $=10 \mathrm{R}$
3 Work done by the gas $=10 \mathrm{R}$
4 Work done $=0$
Explanation:
For 5 moles of gas at temperature T, $\mathrm{PV}_1=5 \mathrm{RT}$ For 5 moles of gas at temperature $(T-2)$, $\mathrm{PV}_2=5 \mathrm{R}(\mathrm{T}-2) \ldots \text { (ii) }$ $\therefore \mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)=5 \mathrm{R}(\mathrm{T}-2-\mathrm{T}) ;$ $\mathrm{P} \Delta \mathrm{V}=-10 \mathrm{R},-\mathrm{P} \Delta \mathrm{V}=10 \mathrm{R}$ When $\Delta \mathrm{V}$ is negative, $\mathrm{W}$ is +ve.
BITSAT-2007
Thermodynamics
272672
Two moles of an ideal gas are compressed isothermally $\left(100^{\circ} \mathrm{C}\right)$ and reversibly from a pressure of $10 \mathrm{~atm}$ to $25 \mathrm{~atm}$. Then the free energy change is:
1 $+15.482 \mathrm{~kJ}$
2 $+10.462 \mathrm{~kJ}$
3 $+5.684 \mathrm{~kJ}$
4 $+3.3642 \mathrm{~kJ}$
Explanation:
For isothermal reversible process: $\mathrm{W}_{\text {rev }} =-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{P}_1}{P_2}\right)=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_1}{\mathrm{~V}_2}$ $=-2.303 \times 2 \times 8.314 \times 373 \times \log \left(\frac{10}{25}\right)=+5684.08 \mathrm{~J}$ $\mathrm{W}_{\text {rev }}=+5.684 \mathrm{~kJ}$ Since, it is a measure of free energy $\therefore$ & $\Delta \mathrm{G}=\mathrm{W}_{\text {rev }}=\mathrm{W}_{\text {max }}$ $\therefore$ & $\Delta \mathrm{G}=+5.684 \mathrm{~kJ}$
AIIMS-2001
Thermodynamics
272673
A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 $\mathrm{dm}^3$ to a volume of $20 \mathrm{dm}^3$. It absorbs $800 \mathrm{~J}$ of thermal energy from its surrounding. The $\Delta \mathrm{U}$ is:
1 $-312 \mathrm{~J}$
2 $+123 \mathrm{~J}$
3 $-213 \mathrm{~J}$
4 $+231 \mathrm{~J}$
Explanation:
In general $\mathrm{W}=-\int_{\mathrm{V}_1}^{\mathrm{V}_2} \mathrm{PdV}=-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)$ $\mathrm{W}=-(1 \mathrm{~atm})(20-10)=-10 \mathrm{dm}^3 \mathrm{~atm}$ $\qquad\left[\begin{array}{l} \because 1 \mathrm{dm}^3=1 \mathrm{~L} \\ 1 \mathrm{atmL}=101.325 \mathrm{~J} \end{array}\right]$ $=-10 \times 101.325=-1013.25 \mathrm{~J} \approx-1013 \mathrm{~J}$ $\text { According to } 1^{\text {st }} \text { law of thermodynamics, }$ $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$ According to $1^{\text {st }}$ law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$
BCECE-2007
Thermodynamics
272675
When the temperature of 2 moles of an ideal gas is increased by $20^{\circ} \mathrm{C}$ at constant pressure. Find the work involved in the process.
272676
The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at $25^{\circ} \mathrm{C}$ is
272683
Temperature of 5 moles of a gas is decreased by $2 \mathrm{~K}$ at constant pressure. Indicate the correct statement
1 Work done by gas is $=5 \mathrm{R}$
2 Work done over the gas is $=10 \mathrm{R}$
3 Work done by the gas $=10 \mathrm{R}$
4 Work done $=0$
Explanation:
For 5 moles of gas at temperature T, $\mathrm{PV}_1=5 \mathrm{RT}$ For 5 moles of gas at temperature $(T-2)$, $\mathrm{PV}_2=5 \mathrm{R}(\mathrm{T}-2) \ldots \text { (ii) }$ $\therefore \mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)=5 \mathrm{R}(\mathrm{T}-2-\mathrm{T}) ;$ $\mathrm{P} \Delta \mathrm{V}=-10 \mathrm{R},-\mathrm{P} \Delta \mathrm{V}=10 \mathrm{R}$ When $\Delta \mathrm{V}$ is negative, $\mathrm{W}$ is +ve.
BITSAT-2007
Thermodynamics
272672
Two moles of an ideal gas are compressed isothermally $\left(100^{\circ} \mathrm{C}\right)$ and reversibly from a pressure of $10 \mathrm{~atm}$ to $25 \mathrm{~atm}$. Then the free energy change is:
1 $+15.482 \mathrm{~kJ}$
2 $+10.462 \mathrm{~kJ}$
3 $+5.684 \mathrm{~kJ}$
4 $+3.3642 \mathrm{~kJ}$
Explanation:
For isothermal reversible process: $\mathrm{W}_{\text {rev }} =-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{P}_1}{P_2}\right)=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_1}{\mathrm{~V}_2}$ $=-2.303 \times 2 \times 8.314 \times 373 \times \log \left(\frac{10}{25}\right)=+5684.08 \mathrm{~J}$ $\mathrm{W}_{\text {rev }}=+5.684 \mathrm{~kJ}$ Since, it is a measure of free energy $\therefore$ & $\Delta \mathrm{G}=\mathrm{W}_{\text {rev }}=\mathrm{W}_{\text {max }}$ $\therefore$ & $\Delta \mathrm{G}=+5.684 \mathrm{~kJ}$
AIIMS-2001
Thermodynamics
272673
A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 $\mathrm{dm}^3$ to a volume of $20 \mathrm{dm}^3$. It absorbs $800 \mathrm{~J}$ of thermal energy from its surrounding. The $\Delta \mathrm{U}$ is:
1 $-312 \mathrm{~J}$
2 $+123 \mathrm{~J}$
3 $-213 \mathrm{~J}$
4 $+231 \mathrm{~J}$
Explanation:
In general $\mathrm{W}=-\int_{\mathrm{V}_1}^{\mathrm{V}_2} \mathrm{PdV}=-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)$ $\mathrm{W}=-(1 \mathrm{~atm})(20-10)=-10 \mathrm{dm}^3 \mathrm{~atm}$ $\qquad\left[\begin{array}{l} \because 1 \mathrm{dm}^3=1 \mathrm{~L} \\ 1 \mathrm{atmL}=101.325 \mathrm{~J} \end{array}\right]$ $=-10 \times 101.325=-1013.25 \mathrm{~J} \approx-1013 \mathrm{~J}$ $\text { According to } 1^{\text {st }} \text { law of thermodynamics, }$ $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$ According to $1^{\text {st }}$ law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$
BCECE-2007
Thermodynamics
272675
When the temperature of 2 moles of an ideal gas is increased by $20^{\circ} \mathrm{C}$ at constant pressure. Find the work involved in the process.
272676
The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at $25^{\circ} \mathrm{C}$ is
272683
Temperature of 5 moles of a gas is decreased by $2 \mathrm{~K}$ at constant pressure. Indicate the correct statement
1 Work done by gas is $=5 \mathrm{R}$
2 Work done over the gas is $=10 \mathrm{R}$
3 Work done by the gas $=10 \mathrm{R}$
4 Work done $=0$
Explanation:
For 5 moles of gas at temperature T, $\mathrm{PV}_1=5 \mathrm{RT}$ For 5 moles of gas at temperature $(T-2)$, $\mathrm{PV}_2=5 \mathrm{R}(\mathrm{T}-2) \ldots \text { (ii) }$ $\therefore \mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)=5 \mathrm{R}(\mathrm{T}-2-\mathrm{T}) ;$ $\mathrm{P} \Delta \mathrm{V}=-10 \mathrm{R},-\mathrm{P} \Delta \mathrm{V}=10 \mathrm{R}$ When $\Delta \mathrm{V}$ is negative, $\mathrm{W}$ is +ve.
BITSAT-2007
Thermodynamics
272672
Two moles of an ideal gas are compressed isothermally $\left(100^{\circ} \mathrm{C}\right)$ and reversibly from a pressure of $10 \mathrm{~atm}$ to $25 \mathrm{~atm}$. Then the free energy change is:
1 $+15.482 \mathrm{~kJ}$
2 $+10.462 \mathrm{~kJ}$
3 $+5.684 \mathrm{~kJ}$
4 $+3.3642 \mathrm{~kJ}$
Explanation:
For isothermal reversible process: $\mathrm{W}_{\text {rev }} =-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{P}_1}{P_2}\right)=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_1}{\mathrm{~V}_2}$ $=-2.303 \times 2 \times 8.314 \times 373 \times \log \left(\frac{10}{25}\right)=+5684.08 \mathrm{~J}$ $\mathrm{W}_{\text {rev }}=+5.684 \mathrm{~kJ}$ Since, it is a measure of free energy $\therefore$ & $\Delta \mathrm{G}=\mathrm{W}_{\text {rev }}=\mathrm{W}_{\text {max }}$ $\therefore$ & $\Delta \mathrm{G}=+5.684 \mathrm{~kJ}$
AIIMS-2001
Thermodynamics
272673
A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 $\mathrm{dm}^3$ to a volume of $20 \mathrm{dm}^3$. It absorbs $800 \mathrm{~J}$ of thermal energy from its surrounding. The $\Delta \mathrm{U}$ is:
1 $-312 \mathrm{~J}$
2 $+123 \mathrm{~J}$
3 $-213 \mathrm{~J}$
4 $+231 \mathrm{~J}$
Explanation:
In general $\mathrm{W}=-\int_{\mathrm{V}_1}^{\mathrm{V}_2} \mathrm{PdV}=-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)$ $\mathrm{W}=-(1 \mathrm{~atm})(20-10)=-10 \mathrm{dm}^3 \mathrm{~atm}$ $\qquad\left[\begin{array}{l} \because 1 \mathrm{dm}^3=1 \mathrm{~L} \\ 1 \mathrm{atmL}=101.325 \mathrm{~J} \end{array}\right]$ $=-10 \times 101.325=-1013.25 \mathrm{~J} \approx-1013 \mathrm{~J}$ $\text { According to } 1^{\text {st }} \text { law of thermodynamics, }$ $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$ According to $1^{\text {st }}$ law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$
BCECE-2007
Thermodynamics
272675
When the temperature of 2 moles of an ideal gas is increased by $20^{\circ} \mathrm{C}$ at constant pressure. Find the work involved in the process.
272676
The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at $25^{\circ} \mathrm{C}$ is
272683
Temperature of 5 moles of a gas is decreased by $2 \mathrm{~K}$ at constant pressure. Indicate the correct statement
1 Work done by gas is $=5 \mathrm{R}$
2 Work done over the gas is $=10 \mathrm{R}$
3 Work done by the gas $=10 \mathrm{R}$
4 Work done $=0$
Explanation:
For 5 moles of gas at temperature T, $\mathrm{PV}_1=5 \mathrm{RT}$ For 5 moles of gas at temperature $(T-2)$, $\mathrm{PV}_2=5 \mathrm{R}(\mathrm{T}-2) \ldots \text { (ii) }$ $\therefore \mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)=5 \mathrm{R}(\mathrm{T}-2-\mathrm{T}) ;$ $\mathrm{P} \Delta \mathrm{V}=-10 \mathrm{R},-\mathrm{P} \Delta \mathrm{V}=10 \mathrm{R}$ When $\Delta \mathrm{V}$ is negative, $\mathrm{W}$ is +ve.
BITSAT-2007
Thermodynamics
272672
Two moles of an ideal gas are compressed isothermally $\left(100^{\circ} \mathrm{C}\right)$ and reversibly from a pressure of $10 \mathrm{~atm}$ to $25 \mathrm{~atm}$. Then the free energy change is:
1 $+15.482 \mathrm{~kJ}$
2 $+10.462 \mathrm{~kJ}$
3 $+5.684 \mathrm{~kJ}$
4 $+3.3642 \mathrm{~kJ}$
Explanation:
For isothermal reversible process: $\mathrm{W}_{\text {rev }} =-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{P}_1}{P_2}\right)=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_1}{\mathrm{~V}_2}$ $=-2.303 \times 2 \times 8.314 \times 373 \times \log \left(\frac{10}{25}\right)=+5684.08 \mathrm{~J}$ $\mathrm{W}_{\text {rev }}=+5.684 \mathrm{~kJ}$ Since, it is a measure of free energy $\therefore$ & $\Delta \mathrm{G}=\mathrm{W}_{\text {rev }}=\mathrm{W}_{\text {max }}$ $\therefore$ & $\Delta \mathrm{G}=+5.684 \mathrm{~kJ}$
AIIMS-2001
Thermodynamics
272673
A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 $\mathrm{dm}^3$ to a volume of $20 \mathrm{dm}^3$. It absorbs $800 \mathrm{~J}$ of thermal energy from its surrounding. The $\Delta \mathrm{U}$ is:
1 $-312 \mathrm{~J}$
2 $+123 \mathrm{~J}$
3 $-213 \mathrm{~J}$
4 $+231 \mathrm{~J}$
Explanation:
In general $\mathrm{W}=-\int_{\mathrm{V}_1}^{\mathrm{V}_2} \mathrm{PdV}=-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)$ $\mathrm{W}=-(1 \mathrm{~atm})(20-10)=-10 \mathrm{dm}^3 \mathrm{~atm}$ $\qquad\left[\begin{array}{l} \because 1 \mathrm{dm}^3=1 \mathrm{~L} \\ 1 \mathrm{atmL}=101.325 \mathrm{~J} \end{array}\right]$ $=-10 \times 101.325=-1013.25 \mathrm{~J} \approx-1013 \mathrm{~J}$ $\text { According to } 1^{\text {st }} \text { law of thermodynamics, }$ $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$ According to $1^{\text {st }}$ law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}=800 \mathrm{~J}+(-1013 \mathrm{~J})=-213 \mathrm{~J}$
BCECE-2007
Thermodynamics
272675
When the temperature of 2 moles of an ideal gas is increased by $20^{\circ} \mathrm{C}$ at constant pressure. Find the work involved in the process.
272676
The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at $25^{\circ} \mathrm{C}$ is
