01. First Law of Thermodynamics and Application
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NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Thermodynamics

272666 One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} \mathrm{C}$. If the work done during the process is $3 \mathrm{~kJ}$, then final temperature of the gas is $\left(\mathrm{C}_{\mathrm{V}}=20 \mathrm{~J} / \mathrm{K}\right)$

1 $100 \mathrm{~K}$
2 $150 \mathrm{~K}$
3 $195 \mathrm{~K}$
4 $255 \mathrm{~K}$
AIIMS-2010
Thermodynamics

272668 6 moles of an ideal gas expand isothermally and reversibly from a volume of 1 litre to a volume of 10 litres at $27^{\circ} \mathrm{C}$. What is the maximum work done?

1 $47 \mathrm{~kJ}$
2 $100 \mathrm{~kJ}$
3 0
4 $34.465 \mathrm{~kJ}$
AIIMS-2009
Thermodynamics

272669 Calculate change in internal energy if $\Delta \mathrm{H}=-92.2 \mathrm{~kJ}, \mathrm{P}=40 \mathrm{~atm}$ and $\Delta \mathrm{V}=-1 \mathrm{~L}$

1 $-42 \mathrm{~kJ}$
2 $-88 \mathrm{~kJ}$
3 $+88 \mathrm{~kJ}$
4 $+42 \mathrm{~kJ}$
AIIMS-2007
Thermodynamics

272670 The enthalpy change $(\triangle H)$ for the reaction, $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})$ is $-92.38 \mathrm{~kJ}$ at 298K. The internal energy change $\Delta \mathrm{U}$ at $298 \mathrm{~K}$ is

1 $-92.38 \mathrm{~kJ}$
2 $-87.42 \mathrm{~kJ}$
3 $-97.34 \mathrm{~kJ}$
4 $-89.9 \mathrm{~kJ}$
AIIMS-2006
NEET DIGITAL LIBRARY
Thermodynamics

272666 One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} \mathrm{C}$. If the work done during the process is $3 \mathrm{~kJ}$, then final temperature of the gas is $\left(\mathrm{C}_{\mathrm{V}}=20 \mathrm{~J} / \mathrm{K}\right)$

1 $100 \mathrm{~K}$
2 $150 \mathrm{~K}$
3 $195 \mathrm{~K}$
4 $255 \mathrm{~K}$
AIIMS-2010
Thermodynamics

272668 6 moles of an ideal gas expand isothermally and reversibly from a volume of 1 litre to a volume of 10 litres at $27^{\circ} \mathrm{C}$. What is the maximum work done?

1 $47 \mathrm{~kJ}$
2 $100 \mathrm{~kJ}$
3 0
4 $34.465 \mathrm{~kJ}$
AIIMS-2009
Thermodynamics

272669 Calculate change in internal energy if $\Delta \mathrm{H}=-92.2 \mathrm{~kJ}, \mathrm{P}=40 \mathrm{~atm}$ and $\Delta \mathrm{V}=-1 \mathrm{~L}$

1 $-42 \mathrm{~kJ}$
2 $-88 \mathrm{~kJ}$
3 $+88 \mathrm{~kJ}$
4 $+42 \mathrm{~kJ}$
AIIMS-2007
Thermodynamics

272670 The enthalpy change $(\triangle H)$ for the reaction, $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})$ is $-92.38 \mathrm{~kJ}$ at 298K. The internal energy change $\Delta \mathrm{U}$ at $298 \mathrm{~K}$ is

1 $-92.38 \mathrm{~kJ}$
2 $-87.42 \mathrm{~kJ}$
3 $-97.34 \mathrm{~kJ}$
4 $-89.9 \mathrm{~kJ}$
AIIMS-2006
Join With Us for More Updates
Thermodynamics

272666 One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} \mathrm{C}$. If the work done during the process is $3 \mathrm{~kJ}$, then final temperature of the gas is $\left(\mathrm{C}_{\mathrm{V}}=20 \mathrm{~J} / \mathrm{K}\right)$

1 $100 \mathrm{~K}$
2 $150 \mathrm{~K}$
3 $195 \mathrm{~K}$
4 $255 \mathrm{~K}$
AIIMS-2010
Thermodynamics

272668 6 moles of an ideal gas expand isothermally and reversibly from a volume of 1 litre to a volume of 10 litres at $27^{\circ} \mathrm{C}$. What is the maximum work done?

1 $47 \mathrm{~kJ}$
2 $100 \mathrm{~kJ}$
3 0
4 $34.465 \mathrm{~kJ}$
AIIMS-2009
Thermodynamics

272669 Calculate change in internal energy if $\Delta \mathrm{H}=-92.2 \mathrm{~kJ}, \mathrm{P}=40 \mathrm{~atm}$ and $\Delta \mathrm{V}=-1 \mathrm{~L}$

1 $-42 \mathrm{~kJ}$
2 $-88 \mathrm{~kJ}$
3 $+88 \mathrm{~kJ}$
4 $+42 \mathrm{~kJ}$
AIIMS-2007
Thermodynamics

272670 The enthalpy change $(\triangle H)$ for the reaction, $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})$ is $-92.38 \mathrm{~kJ}$ at 298K. The internal energy change $\Delta \mathrm{U}$ at $298 \mathrm{~K}$ is

1 $-92.38 \mathrm{~kJ}$
2 $-87.42 \mathrm{~kJ}$
3 $-97.34 \mathrm{~kJ}$
4 $-89.9 \mathrm{~kJ}$
AIIMS-2006
For More Updates join with Us
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Thermodynamics

272666 One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} \mathrm{C}$. If the work done during the process is $3 \mathrm{~kJ}$, then final temperature of the gas is $\left(\mathrm{C}_{\mathrm{V}}=20 \mathrm{~J} / \mathrm{K}\right)$

1 $100 \mathrm{~K}$
2 $150 \mathrm{~K}$
3 $195 \mathrm{~K}$
4 $255 \mathrm{~K}$
AIIMS-2010
Thermodynamics

272668 6 moles of an ideal gas expand isothermally and reversibly from a volume of 1 litre to a volume of 10 litres at $27^{\circ} \mathrm{C}$. What is the maximum work done?

1 $47 \mathrm{~kJ}$
2 $100 \mathrm{~kJ}$
3 0
4 $34.465 \mathrm{~kJ}$
AIIMS-2009
Thermodynamics

272669 Calculate change in internal energy if $\Delta \mathrm{H}=-92.2 \mathrm{~kJ}, \mathrm{P}=40 \mathrm{~atm}$ and $\Delta \mathrm{V}=-1 \mathrm{~L}$

1 $-42 \mathrm{~kJ}$
2 $-88 \mathrm{~kJ}$
3 $+88 \mathrm{~kJ}$
4 $+42 \mathrm{~kJ}$
AIIMS-2007
Thermodynamics

272670 The enthalpy change $(\triangle H)$ for the reaction, $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})$ is $-92.38 \mathrm{~kJ}$ at 298K. The internal energy change $\Delta \mathrm{U}$ at $298 \mathrm{~K}$ is

1 $-92.38 \mathrm{~kJ}$
2 $-87.42 \mathrm{~kJ}$
3 $-97.34 \mathrm{~kJ}$
4 $-89.9 \mathrm{~kJ}$
AIIMS-2006
NEET DIGITAL LIBRARY