148273
The work done by a gas is maximum when it expands :
1 isothermally
2 adiabatically
3 isochorically
4 isobarically
Explanation:
D Work done, $\mathrm{W}=\mathrm{PdV}$. $\mathrm{P}-\mathrm{V}$ diagram of different process is shown below: As work done is the product of pressure and volume changes, so work done is maximum for that process whose curve surrounds maximum area. As shown, area of isobaric curve is maximum hence, work done in isobaric expansion is maximum.
JCECE-2005
Thermodynamics
148277
An ideal gas is expanded from volume $V_{1}$ to volume $V_{2}$, in three different ways: isothermally, adiabatically and isobarically. If $W_{1}, W_{2}$ and $W_{3}$ are respectively the works done in the three processes, then
A Adiabatic, isobaric and isothermal process between $V_{1}$ and $V_{2}$ are shown in the figure. We can see in figure, $\mathrm{A}_{\text {isobaric }}>\mathrm{A}_{\text {isothermal }}>\mathrm{A}_{\text {adiabatic }}$ Hence, $\mathrm{W}_{3}>\mathrm{W}_{2}>\mathrm{W}_{1} \quad[\mathrm{~A}=$ Area, $\mathrm{W}=$ Work done $]$ $\mathrm{A} \propto \mathrm{W}$
COMEDK 2011
Thermodynamics
148278
In an isochoric process
1 Work done is constant.
2 Volume changes, work done remains same
3 Volume remains constant and no work is done by the system
4 Both volume and work done changes
Explanation:
C For an isochoric process the area under the $\mathrm{P}-\mathrm{V}$ curve is zero. Since, no work is done. Thermodynamic process in which the volume remains constant is called isochoric process. $\mathrm{V}=$ constant, $\mathrm{P}=$ variable quantity.
COMEDK 2015
Thermodynamics
148290
During the adiabatic expansion of two moles of a gas the internal energy of a gas is found to decrease by 2 joule. The work done on gas during the process will be equal to
1 $-2 \mathrm{~J}$
2 $3 \mathrm{~J}$
3 $1 \mathrm{~J}$
4 $2 \mathrm{~J}$
Explanation:
A From first law of thermodynamic $\Delta \mathrm{Q}=\Delta \mathrm{W}+\Delta \mathrm{U}$ For adiabatic process, $\mathrm{Q}=0$ $\Delta \mathrm{W}=-\Delta \mathrm{U} \quad(\because \Delta \mathrm{U}=2 \text { Joule })$ $\Delta \mathrm{W}=-2 \text { Joule }$ Hence, work done on gas is -2 Joule.
148273
The work done by a gas is maximum when it expands :
1 isothermally
2 adiabatically
3 isochorically
4 isobarically
Explanation:
D Work done, $\mathrm{W}=\mathrm{PdV}$. $\mathrm{P}-\mathrm{V}$ diagram of different process is shown below: As work done is the product of pressure and volume changes, so work done is maximum for that process whose curve surrounds maximum area. As shown, area of isobaric curve is maximum hence, work done in isobaric expansion is maximum.
JCECE-2005
Thermodynamics
148277
An ideal gas is expanded from volume $V_{1}$ to volume $V_{2}$, in three different ways: isothermally, adiabatically and isobarically. If $W_{1}, W_{2}$ and $W_{3}$ are respectively the works done in the three processes, then
A Adiabatic, isobaric and isothermal process between $V_{1}$ and $V_{2}$ are shown in the figure. We can see in figure, $\mathrm{A}_{\text {isobaric }}>\mathrm{A}_{\text {isothermal }}>\mathrm{A}_{\text {adiabatic }}$ Hence, $\mathrm{W}_{3}>\mathrm{W}_{2}>\mathrm{W}_{1} \quad[\mathrm{~A}=$ Area, $\mathrm{W}=$ Work done $]$ $\mathrm{A} \propto \mathrm{W}$
COMEDK 2011
Thermodynamics
148278
In an isochoric process
1 Work done is constant.
2 Volume changes, work done remains same
3 Volume remains constant and no work is done by the system
4 Both volume and work done changes
Explanation:
C For an isochoric process the area under the $\mathrm{P}-\mathrm{V}$ curve is zero. Since, no work is done. Thermodynamic process in which the volume remains constant is called isochoric process. $\mathrm{V}=$ constant, $\mathrm{P}=$ variable quantity.
COMEDK 2015
Thermodynamics
148290
During the adiabatic expansion of two moles of a gas the internal energy of a gas is found to decrease by 2 joule. The work done on gas during the process will be equal to
1 $-2 \mathrm{~J}$
2 $3 \mathrm{~J}$
3 $1 \mathrm{~J}$
4 $2 \mathrm{~J}$
Explanation:
A From first law of thermodynamic $\Delta \mathrm{Q}=\Delta \mathrm{W}+\Delta \mathrm{U}$ For adiabatic process, $\mathrm{Q}=0$ $\Delta \mathrm{W}=-\Delta \mathrm{U} \quad(\because \Delta \mathrm{U}=2 \text { Joule })$ $\Delta \mathrm{W}=-2 \text { Joule }$ Hence, work done on gas is -2 Joule.
148273
The work done by a gas is maximum when it expands :
1 isothermally
2 adiabatically
3 isochorically
4 isobarically
Explanation:
D Work done, $\mathrm{W}=\mathrm{PdV}$. $\mathrm{P}-\mathrm{V}$ diagram of different process is shown below: As work done is the product of pressure and volume changes, so work done is maximum for that process whose curve surrounds maximum area. As shown, area of isobaric curve is maximum hence, work done in isobaric expansion is maximum.
JCECE-2005
Thermodynamics
148277
An ideal gas is expanded from volume $V_{1}$ to volume $V_{2}$, in three different ways: isothermally, adiabatically and isobarically. If $W_{1}, W_{2}$ and $W_{3}$ are respectively the works done in the three processes, then
A Adiabatic, isobaric and isothermal process between $V_{1}$ and $V_{2}$ are shown in the figure. We can see in figure, $\mathrm{A}_{\text {isobaric }}>\mathrm{A}_{\text {isothermal }}>\mathrm{A}_{\text {adiabatic }}$ Hence, $\mathrm{W}_{3}>\mathrm{W}_{2}>\mathrm{W}_{1} \quad[\mathrm{~A}=$ Area, $\mathrm{W}=$ Work done $]$ $\mathrm{A} \propto \mathrm{W}$
COMEDK 2011
Thermodynamics
148278
In an isochoric process
1 Work done is constant.
2 Volume changes, work done remains same
3 Volume remains constant and no work is done by the system
4 Both volume and work done changes
Explanation:
C For an isochoric process the area under the $\mathrm{P}-\mathrm{V}$ curve is zero. Since, no work is done. Thermodynamic process in which the volume remains constant is called isochoric process. $\mathrm{V}=$ constant, $\mathrm{P}=$ variable quantity.
COMEDK 2015
Thermodynamics
148290
During the adiabatic expansion of two moles of a gas the internal energy of a gas is found to decrease by 2 joule. The work done on gas during the process will be equal to
1 $-2 \mathrm{~J}$
2 $3 \mathrm{~J}$
3 $1 \mathrm{~J}$
4 $2 \mathrm{~J}$
Explanation:
A From first law of thermodynamic $\Delta \mathrm{Q}=\Delta \mathrm{W}+\Delta \mathrm{U}$ For adiabatic process, $\mathrm{Q}=0$ $\Delta \mathrm{W}=-\Delta \mathrm{U} \quad(\because \Delta \mathrm{U}=2 \text { Joule })$ $\Delta \mathrm{W}=-2 \text { Joule }$ Hence, work done on gas is -2 Joule.
148273
The work done by a gas is maximum when it expands :
1 isothermally
2 adiabatically
3 isochorically
4 isobarically
Explanation:
D Work done, $\mathrm{W}=\mathrm{PdV}$. $\mathrm{P}-\mathrm{V}$ diagram of different process is shown below: As work done is the product of pressure and volume changes, so work done is maximum for that process whose curve surrounds maximum area. As shown, area of isobaric curve is maximum hence, work done in isobaric expansion is maximum.
JCECE-2005
Thermodynamics
148277
An ideal gas is expanded from volume $V_{1}$ to volume $V_{2}$, in three different ways: isothermally, adiabatically and isobarically. If $W_{1}, W_{2}$ and $W_{3}$ are respectively the works done in the three processes, then
A Adiabatic, isobaric and isothermal process between $V_{1}$ and $V_{2}$ are shown in the figure. We can see in figure, $\mathrm{A}_{\text {isobaric }}>\mathrm{A}_{\text {isothermal }}>\mathrm{A}_{\text {adiabatic }}$ Hence, $\mathrm{W}_{3}>\mathrm{W}_{2}>\mathrm{W}_{1} \quad[\mathrm{~A}=$ Area, $\mathrm{W}=$ Work done $]$ $\mathrm{A} \propto \mathrm{W}$
COMEDK 2011
Thermodynamics
148278
In an isochoric process
1 Work done is constant.
2 Volume changes, work done remains same
3 Volume remains constant and no work is done by the system
4 Both volume and work done changes
Explanation:
C For an isochoric process the area under the $\mathrm{P}-\mathrm{V}$ curve is zero. Since, no work is done. Thermodynamic process in which the volume remains constant is called isochoric process. $\mathrm{V}=$ constant, $\mathrm{P}=$ variable quantity.
COMEDK 2015
Thermodynamics
148290
During the adiabatic expansion of two moles of a gas the internal energy of a gas is found to decrease by 2 joule. The work done on gas during the process will be equal to
1 $-2 \mathrm{~J}$
2 $3 \mathrm{~J}$
3 $1 \mathrm{~J}$
4 $2 \mathrm{~J}$
Explanation:
A From first law of thermodynamic $\Delta \mathrm{Q}=\Delta \mathrm{W}+\Delta \mathrm{U}$ For adiabatic process, $\mathrm{Q}=0$ $\Delta \mathrm{W}=-\Delta \mathrm{U} \quad(\because \Delta \mathrm{U}=2 \text { Joule })$ $\Delta \mathrm{W}=-2 \text { Joule }$ Hence, work done on gas is -2 Joule.
