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Thermodynamics

148389 In isochoric process

1

Δ W = 0

2

Δ U = 0

3

Δ Q = 0

4

Δ T = 0

Explanation:

A - The work done by an ideal gas during an isochoric process is zero.
- In Isochoric process, volume remains constant. Some energy with its environment, the process is known as an isochoric process.
So, When volume $(V) =$ Constant
Then $dV = 0$
and
work done $(W) = \int PdV = 0$
It mean $W = 0$

AP EAMCET-25.08.2021

Thermodynamics

148390 A process in which the amount of heat supplied to the system goes fully to change its internal energy and temperature is

1 adiabatic process

2 cyclic process

3 isobaric process

4 isothermal process

5 isochoric process

Explanation:

E According to the first law of thermodynamics, $Δ Q = Δ W + Δ U$
In an isochoric process, the volume remains constant $(dV = 0)$
So work done $(Δ W) = PdV = 0$
Hence, $Δ Q = 0 + Δ U$
$Δ Q = Δ U$
Heat exchanged $=$ change in internal energy.

Kerala CEE 04.07.2022

Thermodynamics

148391 The Volume of $1 m^{3}$ of gas is doubled at atmospheric pressure. The work done at constant pressure will be

1 zero

2

10^{5} cal

3

10^{5} J

4

10^{5} erg

[SRM JEE-2018]

Explanation:

C Isobaric process- An isobaric process is a process occurring at constant pressure.
For isobaric process we write $Δ U = Δ Q - Δ W$
$= Δ Q - P (V_{2} - V_{1})$
Work done in isobaric process is given as $W$
$= P (V_{f} - V_{i})$
$= 1 \times 10^{5} (2 - 1)$
$= 10^{5} J$

Thermodynamics

148392 A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by ${20.0}^{\circ} C$ will be
(Given gas constant $R = 8.3 {JK}^{- 1}$ -mol $^{- 1}$ )

1

249 J

2

415 J

3

498 J

4

830 J

Explanation:

C Given, Volume $= 44.8$ lit $= 44.8 \times 10^{- 3} m^{3}$ We know that,
One mole of any ideal gas at STP occupies a volume of 22.4 litre therefore, cylinder of fixed capacity 44.8 litre must contain 2 moles of helium at STP.
As helium is monatomic gas.
$For helium, C_{v} = \frac{3}{2} R$
Now, Heat needed to raise the temperature
$Q = {nC}_{v} dT$
$Q = 2 \times \frac{3}{2} R \times 20$
$= 3 \times 8.3 \times 20 = 498 J$

JEE Main-29.06.2022

Thermodynamics

148393 In an isochoric process, if heat is supplied to gas the pressure will

1 Increase

2 Decrease

3 Remains same

4 Varies exponentially

Explanation:

A From ideal gas equation $PV = nRT$
$∴$ For Isochoric process $(V = C)$ ,
P $\propto T$
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
For isochoric process temperature is directly
proportional to pressure. If we supply heat, temperature
increases so pressure also increases.

HP CET-2018

Thermodynamics

148389 In isochoric process

1

Δ W = 0

2

Δ U = 0

3

Δ Q = 0

4

Δ T = 0

Explanation:

A - The work done by an ideal gas during an isochoric process is zero.
- In Isochoric process, volume remains constant. Some energy with its environment, the process is known as an isochoric process.
So, When volume $(V) =$ Constant
Then $dV = 0$
and
work done $(W) = \int PdV = 0$
It mean $W = 0$

AP EAMCET-25.08.2021

Thermodynamics

148390 A process in which the amount of heat supplied to the system goes fully to change its internal energy and temperature is

1 adiabatic process

2 cyclic process

3 isobaric process

4 isothermal process

5 isochoric process

Explanation:

E According to the first law of thermodynamics, $Δ Q = Δ W + Δ U$
In an isochoric process, the volume remains constant $(dV = 0)$
So work done $(Δ W) = PdV = 0$
Hence, $Δ Q = 0 + Δ U$
$Δ Q = Δ U$
Heat exchanged $=$ change in internal energy.

Kerala CEE 04.07.2022

Thermodynamics

148391 The Volume of $1 m^{3}$ of gas is doubled at atmospheric pressure. The work done at constant pressure will be

1 zero

2

10^{5} cal

3

10^{5} J

4

10^{5} erg

[SRM JEE-2018]

Explanation:

C Isobaric process- An isobaric process is a process occurring at constant pressure.
For isobaric process we write $Δ U = Δ Q - Δ W$
$= Δ Q - P (V_{2} - V_{1})$
Work done in isobaric process is given as $W$
$= P (V_{f} - V_{i})$
$= 1 \times 10^{5} (2 - 1)$
$= 10^{5} J$

Thermodynamics

148392 A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by ${20.0}^{\circ} C$ will be
(Given gas constant $R = 8.3 {JK}^{- 1}$ -mol $^{- 1}$ )

1

249 J

2

415 J

3

498 J

4

830 J

Explanation:

C Given, Volume $= 44.8$ lit $= 44.8 \times 10^{- 3} m^{3}$ We know that,
One mole of any ideal gas at STP occupies a volume of 22.4 litre therefore, cylinder of fixed capacity 44.8 litre must contain 2 moles of helium at STP.
As helium is monatomic gas.
$For helium, C_{v} = \frac{3}{2} R$
Now, Heat needed to raise the temperature
$Q = {nC}_{v} dT$
$Q = 2 \times \frac{3}{2} R \times 20$
$= 3 \times 8.3 \times 20 = 498 J$

JEE Main-29.06.2022

Thermodynamics

148393 In an isochoric process, if heat is supplied to gas the pressure will

1 Increase

2 Decrease

3 Remains same

4 Varies exponentially

Explanation:

A From ideal gas equation $PV = nRT$
$∴$ For Isochoric process $(V = C)$ ,
P $\propto T$
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
For isochoric process temperature is directly
proportional to pressure. If we supply heat, temperature
increases so pressure also increases.

HP CET-2018

Thermodynamics

148389 In isochoric process

1

Δ W = 0

2

Δ U = 0

3

Δ Q = 0

4

Δ T = 0

Explanation:

A - The work done by an ideal gas during an isochoric process is zero.
- In Isochoric process, volume remains constant. Some energy with its environment, the process is known as an isochoric process.
So, When volume $(V) =$ Constant
Then $dV = 0$
and
work done $(W) = \int PdV = 0$
It mean $W = 0$

AP EAMCET-25.08.2021

Thermodynamics

148390 A process in which the amount of heat supplied to the system goes fully to change its internal energy and temperature is

1 adiabatic process

2 cyclic process

3 isobaric process

4 isothermal process

5 isochoric process

Explanation:

E According to the first law of thermodynamics, $Δ Q = Δ W + Δ U$
In an isochoric process, the volume remains constant $(dV = 0)$
So work done $(Δ W) = PdV = 0$
Hence, $Δ Q = 0 + Δ U$
$Δ Q = Δ U$
Heat exchanged $=$ change in internal energy.

Kerala CEE 04.07.2022

Thermodynamics

148391 The Volume of $1 m^{3}$ of gas is doubled at atmospheric pressure. The work done at constant pressure will be

1 zero

2

10^{5} cal

3

10^{5} J

4

10^{5} erg

[SRM JEE-2018]

Explanation:

C Isobaric process- An isobaric process is a process occurring at constant pressure.
For isobaric process we write $Δ U = Δ Q - Δ W$
$= Δ Q - P (V_{2} - V_{1})$
Work done in isobaric process is given as $W$
$= P (V_{f} - V_{i})$
$= 1 \times 10^{5} (2 - 1)$
$= 10^{5} J$

Thermodynamics

148392 A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by ${20.0}^{\circ} C$ will be
(Given gas constant $R = 8.3 {JK}^{- 1}$ -mol $^{- 1}$ )

1

249 J

2

415 J

3

498 J

4

830 J

Explanation:

C Given, Volume $= 44.8$ lit $= 44.8 \times 10^{- 3} m^{3}$ We know that,
One mole of any ideal gas at STP occupies a volume of 22.4 litre therefore, cylinder of fixed capacity 44.8 litre must contain 2 moles of helium at STP.
As helium is monatomic gas.
$For helium, C_{v} = \frac{3}{2} R$
Now, Heat needed to raise the temperature
$Q = {nC}_{v} dT$
$Q = 2 \times \frac{3}{2} R \times 20$
$= 3 \times 8.3 \times 20 = 498 J$

JEE Main-29.06.2022

Thermodynamics

148393 In an isochoric process, if heat is supplied to gas the pressure will

1 Increase

2 Decrease

3 Remains same

4 Varies exponentially

Explanation:

A From ideal gas equation $PV = nRT$
$∴$ For Isochoric process $(V = C)$ ,
P $\propto T$
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
For isochoric process temperature is directly
proportional to pressure. If we supply heat, temperature
increases so pressure also increases.

HP CET-2018

Thermodynamics

148389 In isochoric process

1

Δ W = 0

2

Δ U = 0

3

Δ Q = 0

4

Δ T = 0

Explanation:

A - The work done by an ideal gas during an isochoric process is zero.
- In Isochoric process, volume remains constant. Some energy with its environment, the process is known as an isochoric process.
So, When volume $(V) =$ Constant
Then $dV = 0$
and
work done $(W) = \int PdV = 0$
It mean $W = 0$

AP EAMCET-25.08.2021

Thermodynamics

148390 A process in which the amount of heat supplied to the system goes fully to change its internal energy and temperature is

1 adiabatic process

2 cyclic process

3 isobaric process

4 isothermal process

5 isochoric process

Explanation:

E According to the first law of thermodynamics, $Δ Q = Δ W + Δ U$
In an isochoric process, the volume remains constant $(dV = 0)$
So work done $(Δ W) = PdV = 0$
Hence, $Δ Q = 0 + Δ U$
$Δ Q = Δ U$
Heat exchanged $=$ change in internal energy.

Kerala CEE 04.07.2022

Thermodynamics

148391 The Volume of $1 m^{3}$ of gas is doubled at atmospheric pressure. The work done at constant pressure will be

1 zero

2

10^{5} cal

3

10^{5} J

4

10^{5} erg

[SRM JEE-2018]

Explanation:

C Isobaric process- An isobaric process is a process occurring at constant pressure.
For isobaric process we write $Δ U = Δ Q - Δ W$
$= Δ Q - P (V_{2} - V_{1})$
Work done in isobaric process is given as $W$
$= P (V_{f} - V_{i})$
$= 1 \times 10^{5} (2 - 1)$
$= 10^{5} J$

Thermodynamics

148392 A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by ${20.0}^{\circ} C$ will be
(Given gas constant $R = 8.3 {JK}^{- 1}$ -mol $^{- 1}$ )

1

249 J

2

415 J

3

498 J

4

830 J

Explanation:

C Given, Volume $= 44.8$ lit $= 44.8 \times 10^{- 3} m^{3}$ We know that,
One mole of any ideal gas at STP occupies a volume of 22.4 litre therefore, cylinder of fixed capacity 44.8 litre must contain 2 moles of helium at STP.
As helium is monatomic gas.
$For helium, C_{v} = \frac{3}{2} R$
Now, Heat needed to raise the temperature
$Q = {nC}_{v} dT$
$Q = 2 \times \frac{3}{2} R \times 20$
$= 3 \times 8.3 \times 20 = 498 J$

JEE Main-29.06.2022

Thermodynamics

148393 In an isochoric process, if heat is supplied to gas the pressure will

1 Increase

2 Decrease

3 Remains same

4 Varies exponentially

Explanation:

A From ideal gas equation $PV = nRT$
$∴$ For Isochoric process $(V = C)$ ,
P $\propto T$
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
For isochoric process temperature is directly
proportional to pressure. If we supply heat, temperature
increases so pressure also increases.

HP CET-2018

Thermodynamics

148389 In isochoric process

1

Δ W = 0

2

Δ U = 0

3

Δ Q = 0

4

Δ T = 0

Explanation:

A - The work done by an ideal gas during an isochoric process is zero.
- In Isochoric process, volume remains constant. Some energy with its environment, the process is known as an isochoric process.
So, When volume $(V) =$ Constant
Then $dV = 0$
and
work done $(W) = \int PdV = 0$
It mean $W = 0$

AP EAMCET-25.08.2021

Thermodynamics

148390 A process in which the amount of heat supplied to the system goes fully to change its internal energy and temperature is

1 adiabatic process

2 cyclic process

3 isobaric process

4 isothermal process

5 isochoric process

Explanation:

E According to the first law of thermodynamics, $Δ Q = Δ W + Δ U$
In an isochoric process, the volume remains constant $(dV = 0)$
So work done $(Δ W) = PdV = 0$
Hence, $Δ Q = 0 + Δ U$
$Δ Q = Δ U$
Heat exchanged $=$ change in internal energy.

Kerala CEE 04.07.2022

Thermodynamics

148391 The Volume of $1 m^{3}$ of gas is doubled at atmospheric pressure. The work done at constant pressure will be

1 zero

2

10^{5} cal

3

10^{5} J

4

10^{5} erg

[SRM JEE-2018]

Explanation:

C Isobaric process- An isobaric process is a process occurring at constant pressure.
For isobaric process we write $Δ U = Δ Q - Δ W$
$= Δ Q - P (V_{2} - V_{1})$
Work done in isobaric process is given as $W$
$= P (V_{f} - V_{i})$
$= 1 \times 10^{5} (2 - 1)$
$= 10^{5} J$

Thermodynamics

148392 A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by ${20.0}^{\circ} C$ will be
(Given gas constant $R = 8.3 {JK}^{- 1}$ -mol $^{- 1}$ )

1

249 J

2

415 J

3

498 J

4

830 J

Explanation:

C Given, Volume $= 44.8$ lit $= 44.8 \times 10^{- 3} m^{3}$ We know that,
One mole of any ideal gas at STP occupies a volume of 22.4 litre therefore, cylinder of fixed capacity 44.8 litre must contain 2 moles of helium at STP.
As helium is monatomic gas.
$For helium, C_{v} = \frac{3}{2} R$
Now, Heat needed to raise the temperature
$Q = {nC}_{v} dT$
$Q = 2 \times \frac{3}{2} R \times 20$
$= 3 \times 8.3 \times 20 = 498 J$

JEE Main-29.06.2022

Thermodynamics

148393 In an isochoric process, if heat is supplied to gas the pressure will

1 Increase

2 Decrease

3 Remains same

4 Varies exponentially

Explanation:

A From ideal gas equation $PV = nRT$
$∴$ For Isochoric process $(V = C)$ ,
P $\propto T$
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
For isochoric process temperature is directly
proportional to pressure. If we supply heat, temperature
increases so pressure also increases.

HP CET-2018