369326
What is the change in the energy of system if \(500 \mathrm{cal}\) of heat energy are added to a system and system does \(350 \mathrm{cal}\) of work on the surroundings.
1 \(-150 \mathrm{cal}\)
2 \(+150 \mathrm{cal}\)
3 \(+850 \mathrm{cal}\)
4 \(-850 \mathrm{cal}\)
Explanation:
Heat absorbed, \({\text{q = 500}}{\mkern 1mu} {\text{cal}}\) Work done by the system, \({\text{w = - 350}}\,{\text{cal}}\) According to the first law of thermodynamics, \(\Delta \mathrm{E}=\mathrm{q}+\mathrm{w}=500+(-350)=+150 \mathrm{cal}\)
AIIMS - 2019
CHXI06:THERMODYNAMICS
369327
A gas is allowed to expand in an insulated container against a constant external pressure of \(2.5 \mathrm{~atm}\) form \(2.5 \mathrm{~L}\) to \(4.5 \mathrm{~L}\), the change in internal energy of the gas in joules is
1 \(-836.3 \mathrm{~J}\)
2 \(-1136.2 \mathrm{~J}\)
3 \(-450 \mathrm{~J}\)
4 \(-506.5 \mathrm{~J}\)
Explanation:
The change in internal energy is given by, \(\Delta \mathrm{U}=\mathrm{Q}+\mathrm{W}\) For insulated container, \(\mathrm{Q}=0\) \(\therefore \Delta \mathrm{U}=\mathrm{W}=-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\text {ex }}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)\) \(\therefore \Delta \mathrm{U}=-2.5 \mathrm{~atm}(4.5 \mathrm{~L}-2.5 \mathrm{~L})\) \(=-5.0 \mathrm{~L} \cdot \mathrm{atm}\) \(=-5.0 \times 101.325 \mathrm{~J} \quad(1 \mathrm{~L} \cdot \mathrm{atm}=101.325 \mathrm{~J})\) \(\therefore \Delta \mathrm{U}=-506.625 \mathrm{~J}\)
MHTCET - 2021
CHXI06:THERMODYNAMICS
369328
When \(\mathrm{1 \mathrm{~mol}}\) of a gas is heated at constant volume, temperature is raised from 298 to 308 \(\mathrm{\mathrm{K}}\). If heat supplied to the gas is \(\mathrm{500 \mathrm{~J}}\), then which statement is correct?
As volume is constant hence work done in this process is zero therefore heat supplied is equal to change in internal energy. \(\therefore \,\,{\rm{\Delta V}}\,\,{\rm{ = 0;}}\,\,{\rm{w = 0}}\) According to \({{\rm{1}}^{{\rm{st}}}}\) law of thermodynamics, \({\rm{\Delta U = q + w = q = 500}}\,{\rm{J}}{\rm{.}}\)
CHXI06:THERMODYNAMICS
369329
A gas performs \(\mathrm{0.320 \mathrm{~kJ}}\) work on surrounding and absorbs \(\mathrm{120 \mathrm{~J}}\) of heat from the surrounding. Hence, change in internal energy is
1 \(\mathrm{200 \mathrm{~J}}\)
2 \(\mathrm{120.32 \mathrm{~J}}\)
3 \(\mathrm{-200 \mathrm{~J}}\)
4 \(\mathrm{440 \mathrm{~J}}\)
Explanation:
According to first law of thermodynamics, \(\mathrm{\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}}\) Since, work is done on the surrounding. So, \(\mathrm{\quad \mathrm{W}=-0.320 \mathrm{~kJ}}\) \({\rm{ = - 320}}\;{\rm{J}}\,{\rm{q = 120}}\;{\rm{J}}\) \(\therefore \Delta {\rm{U = - 320 + 120 = - 200}}\;{\rm{J}}\)
369326
What is the change in the energy of system if \(500 \mathrm{cal}\) of heat energy are added to a system and system does \(350 \mathrm{cal}\) of work on the surroundings.
1 \(-150 \mathrm{cal}\)
2 \(+150 \mathrm{cal}\)
3 \(+850 \mathrm{cal}\)
4 \(-850 \mathrm{cal}\)
Explanation:
Heat absorbed, \({\text{q = 500}}{\mkern 1mu} {\text{cal}}\) Work done by the system, \({\text{w = - 350}}\,{\text{cal}}\) According to the first law of thermodynamics, \(\Delta \mathrm{E}=\mathrm{q}+\mathrm{w}=500+(-350)=+150 \mathrm{cal}\)
AIIMS - 2019
CHXI06:THERMODYNAMICS
369327
A gas is allowed to expand in an insulated container against a constant external pressure of \(2.5 \mathrm{~atm}\) form \(2.5 \mathrm{~L}\) to \(4.5 \mathrm{~L}\), the change in internal energy of the gas in joules is
1 \(-836.3 \mathrm{~J}\)
2 \(-1136.2 \mathrm{~J}\)
3 \(-450 \mathrm{~J}\)
4 \(-506.5 \mathrm{~J}\)
Explanation:
The change in internal energy is given by, \(\Delta \mathrm{U}=\mathrm{Q}+\mathrm{W}\) For insulated container, \(\mathrm{Q}=0\) \(\therefore \Delta \mathrm{U}=\mathrm{W}=-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\text {ex }}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)\) \(\therefore \Delta \mathrm{U}=-2.5 \mathrm{~atm}(4.5 \mathrm{~L}-2.5 \mathrm{~L})\) \(=-5.0 \mathrm{~L} \cdot \mathrm{atm}\) \(=-5.0 \times 101.325 \mathrm{~J} \quad(1 \mathrm{~L} \cdot \mathrm{atm}=101.325 \mathrm{~J})\) \(\therefore \Delta \mathrm{U}=-506.625 \mathrm{~J}\)
MHTCET - 2021
CHXI06:THERMODYNAMICS
369328
When \(\mathrm{1 \mathrm{~mol}}\) of a gas is heated at constant volume, temperature is raised from 298 to 308 \(\mathrm{\mathrm{K}}\). If heat supplied to the gas is \(\mathrm{500 \mathrm{~J}}\), then which statement is correct?
As volume is constant hence work done in this process is zero therefore heat supplied is equal to change in internal energy. \(\therefore \,\,{\rm{\Delta V}}\,\,{\rm{ = 0;}}\,\,{\rm{w = 0}}\) According to \({{\rm{1}}^{{\rm{st}}}}\) law of thermodynamics, \({\rm{\Delta U = q + w = q = 500}}\,{\rm{J}}{\rm{.}}\)
CHXI06:THERMODYNAMICS
369329
A gas performs \(\mathrm{0.320 \mathrm{~kJ}}\) work on surrounding and absorbs \(\mathrm{120 \mathrm{~J}}\) of heat from the surrounding. Hence, change in internal energy is
1 \(\mathrm{200 \mathrm{~J}}\)
2 \(\mathrm{120.32 \mathrm{~J}}\)
3 \(\mathrm{-200 \mathrm{~J}}\)
4 \(\mathrm{440 \mathrm{~J}}\)
Explanation:
According to first law of thermodynamics, \(\mathrm{\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}}\) Since, work is done on the surrounding. So, \(\mathrm{\quad \mathrm{W}=-0.320 \mathrm{~kJ}}\) \({\rm{ = - 320}}\;{\rm{J}}\,{\rm{q = 120}}\;{\rm{J}}\) \(\therefore \Delta {\rm{U = - 320 + 120 = - 200}}\;{\rm{J}}\)
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CHXI06:THERMODYNAMICS
369326
What is the change in the energy of system if \(500 \mathrm{cal}\) of heat energy are added to a system and system does \(350 \mathrm{cal}\) of work on the surroundings.
1 \(-150 \mathrm{cal}\)
2 \(+150 \mathrm{cal}\)
3 \(+850 \mathrm{cal}\)
4 \(-850 \mathrm{cal}\)
Explanation:
Heat absorbed, \({\text{q = 500}}{\mkern 1mu} {\text{cal}}\) Work done by the system, \({\text{w = - 350}}\,{\text{cal}}\) According to the first law of thermodynamics, \(\Delta \mathrm{E}=\mathrm{q}+\mathrm{w}=500+(-350)=+150 \mathrm{cal}\)
AIIMS - 2019
CHXI06:THERMODYNAMICS
369327
A gas is allowed to expand in an insulated container against a constant external pressure of \(2.5 \mathrm{~atm}\) form \(2.5 \mathrm{~L}\) to \(4.5 \mathrm{~L}\), the change in internal energy of the gas in joules is
1 \(-836.3 \mathrm{~J}\)
2 \(-1136.2 \mathrm{~J}\)
3 \(-450 \mathrm{~J}\)
4 \(-506.5 \mathrm{~J}\)
Explanation:
The change in internal energy is given by, \(\Delta \mathrm{U}=\mathrm{Q}+\mathrm{W}\) For insulated container, \(\mathrm{Q}=0\) \(\therefore \Delta \mathrm{U}=\mathrm{W}=-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\text {ex }}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)\) \(\therefore \Delta \mathrm{U}=-2.5 \mathrm{~atm}(4.5 \mathrm{~L}-2.5 \mathrm{~L})\) \(=-5.0 \mathrm{~L} \cdot \mathrm{atm}\) \(=-5.0 \times 101.325 \mathrm{~J} \quad(1 \mathrm{~L} \cdot \mathrm{atm}=101.325 \mathrm{~J})\) \(\therefore \Delta \mathrm{U}=-506.625 \mathrm{~J}\)
MHTCET - 2021
CHXI06:THERMODYNAMICS
369328
When \(\mathrm{1 \mathrm{~mol}}\) of a gas is heated at constant volume, temperature is raised from 298 to 308 \(\mathrm{\mathrm{K}}\). If heat supplied to the gas is \(\mathrm{500 \mathrm{~J}}\), then which statement is correct?
As volume is constant hence work done in this process is zero therefore heat supplied is equal to change in internal energy. \(\therefore \,\,{\rm{\Delta V}}\,\,{\rm{ = 0;}}\,\,{\rm{w = 0}}\) According to \({{\rm{1}}^{{\rm{st}}}}\) law of thermodynamics, \({\rm{\Delta U = q + w = q = 500}}\,{\rm{J}}{\rm{.}}\)
CHXI06:THERMODYNAMICS
369329
A gas performs \(\mathrm{0.320 \mathrm{~kJ}}\) work on surrounding and absorbs \(\mathrm{120 \mathrm{~J}}\) of heat from the surrounding. Hence, change in internal energy is
1 \(\mathrm{200 \mathrm{~J}}\)
2 \(\mathrm{120.32 \mathrm{~J}}\)
3 \(\mathrm{-200 \mathrm{~J}}\)
4 \(\mathrm{440 \mathrm{~J}}\)
Explanation:
According to first law of thermodynamics, \(\mathrm{\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}}\) Since, work is done on the surrounding. So, \(\mathrm{\quad \mathrm{W}=-0.320 \mathrm{~kJ}}\) \({\rm{ = - 320}}\;{\rm{J}}\,{\rm{q = 120}}\;{\rm{J}}\) \(\therefore \Delta {\rm{U = - 320 + 120 = - 200}}\;{\rm{J}}\)
369326
What is the change in the energy of system if \(500 \mathrm{cal}\) of heat energy are added to a system and system does \(350 \mathrm{cal}\) of work on the surroundings.
1 \(-150 \mathrm{cal}\)
2 \(+150 \mathrm{cal}\)
3 \(+850 \mathrm{cal}\)
4 \(-850 \mathrm{cal}\)
Explanation:
Heat absorbed, \({\text{q = 500}}{\mkern 1mu} {\text{cal}}\) Work done by the system, \({\text{w = - 350}}\,{\text{cal}}\) According to the first law of thermodynamics, \(\Delta \mathrm{E}=\mathrm{q}+\mathrm{w}=500+(-350)=+150 \mathrm{cal}\)
AIIMS - 2019
CHXI06:THERMODYNAMICS
369327
A gas is allowed to expand in an insulated container against a constant external pressure of \(2.5 \mathrm{~atm}\) form \(2.5 \mathrm{~L}\) to \(4.5 \mathrm{~L}\), the change in internal energy of the gas in joules is
1 \(-836.3 \mathrm{~J}\)
2 \(-1136.2 \mathrm{~J}\)
3 \(-450 \mathrm{~J}\)
4 \(-506.5 \mathrm{~J}\)
Explanation:
The change in internal energy is given by, \(\Delta \mathrm{U}=\mathrm{Q}+\mathrm{W}\) For insulated container, \(\mathrm{Q}=0\) \(\therefore \Delta \mathrm{U}=\mathrm{W}=-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\text {ex }}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)\) \(\therefore \Delta \mathrm{U}=-2.5 \mathrm{~atm}(4.5 \mathrm{~L}-2.5 \mathrm{~L})\) \(=-5.0 \mathrm{~L} \cdot \mathrm{atm}\) \(=-5.0 \times 101.325 \mathrm{~J} \quad(1 \mathrm{~L} \cdot \mathrm{atm}=101.325 \mathrm{~J})\) \(\therefore \Delta \mathrm{U}=-506.625 \mathrm{~J}\)
MHTCET - 2021
CHXI06:THERMODYNAMICS
369328
When \(\mathrm{1 \mathrm{~mol}}\) of a gas is heated at constant volume, temperature is raised from 298 to 308 \(\mathrm{\mathrm{K}}\). If heat supplied to the gas is \(\mathrm{500 \mathrm{~J}}\), then which statement is correct?
As volume is constant hence work done in this process is zero therefore heat supplied is equal to change in internal energy. \(\therefore \,\,{\rm{\Delta V}}\,\,{\rm{ = 0;}}\,\,{\rm{w = 0}}\) According to \({{\rm{1}}^{{\rm{st}}}}\) law of thermodynamics, \({\rm{\Delta U = q + w = q = 500}}\,{\rm{J}}{\rm{.}}\)
CHXI06:THERMODYNAMICS
369329
A gas performs \(\mathrm{0.320 \mathrm{~kJ}}\) work on surrounding and absorbs \(\mathrm{120 \mathrm{~J}}\) of heat from the surrounding. Hence, change in internal energy is
1 \(\mathrm{200 \mathrm{~J}}\)
2 \(\mathrm{120.32 \mathrm{~J}}\)
3 \(\mathrm{-200 \mathrm{~J}}\)
4 \(\mathrm{440 \mathrm{~J}}\)
Explanation:
According to first law of thermodynamics, \(\mathrm{\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}}\) Since, work is done on the surrounding. So, \(\mathrm{\quad \mathrm{W}=-0.320 \mathrm{~kJ}}\) \({\rm{ = - 320}}\;{\rm{J}}\,{\rm{q = 120}}\;{\rm{J}}\) \(\therefore \Delta {\rm{U = - 320 + 120 = - 200}}\;{\rm{J}}\)