QBManager

Thermodynamics

148422 When $50 l$ of air at STP (Standard Temperature and Pressure) is compressed isothermally to $10 l$ , the amount and direction of heat exchange of air is
$[Use P_{atm} = 100 kPa and \ln (0.2) = - 1.61]$

1

8.05 kJ

flows out

2

8.05 kJ

flows in

3

5 kJ

flows out

4

5 kJ

flows in

Explanation:

A Given, Initial volume of air at $STP = 50 l = 50 \times 10^{- 3} m^{3}$ Final Volume of air $= 10 l = 10 \times 10^{- 3} m^{3}$
$P_{atm} = 100 kPa$
Now, Process is Isothermally compressed
$∴$ From $1^{st}$ law of thermodynamics
$Δ Q = Δ U + Δ W$
$∴ Δ Q = Δ W$
Now, for Isothermal process
$Δ W = P_{i} V_{i} \ln (\frac{V_{f}}{V_{i}})$
$= 100 \times 50 \times 10^{- 3} \times \ln (\frac{10 \times 10^{- 3}}{50 \times 10^{- 3}})$
$= 5 \times \ln (0.2)$
$= - 5 \times 1.61$
$Δ Q = - 8.05 kJ$
Negative sign shows that heat rejected from the system
$∴ Δ Q = 8.05 kJ flows out$

TS EAMCET 02.05.2018

Thermodynamics

148424 A graph of pressure (P) against volume (V) of an ideal gas undergoing an isothermal process is:

1 a straight line passing through the origin

2 a parabola

3 a rectangular hyperbola

4 a straight line parallel to pressure axis

Explanation:

C We know that

Ideal gas equation
$PV = nRT$ .....(i)
For isothermal process
$PV =$ Constant
Now, Differentiating both side
$PdV + VdP = 0$
$\frac{dP}{dV} = - \frac{V}{P}$
Note: differential equation of rectangular hyperbola.

AP EAMCET-23.09.2020

Thermodynamics

148425 An ideal gas expands isothermally from volume $V_{1}$ to volume $V_{2}$ . It is then compressed to the original volume $V_{1}$ adiabatically. If $p_{1}, p_{2}$ and $W$ represent the initial pressure, final pressure and the net work done by the gas respectively during the entire process, then

1

p_{1} > p_{2}, W = 0

2

p_{1} > p_{2}, W > 0

3

p_{2} > p_{1}, W > 0

4

p_{2} > p_{1}, W < 0

Explanation:

D Given,

Isothermal
Adiabatic
Expansion
$T_{A} = T_{B}$
Now,
As shown in the above figure
Process $AB =$ Isothermal Expansion
Process $BC =$ Adiabatic Compression From figure it can be said that, area under the curve for isothermal expansion is lesser than area under the curve for adiabatic compression.
$∴$ Work done is negative $(W < 0)$ . Also pressure $(P_{2} > P_{1})$ .

AP EAMCET - 2010

Thermodynamics

1 A iv , B ii , C i , D iii

2 A iv , B i , C ii , D iii

3 A ii , B iii , C iv , D i

4 A ii , B i , C iv , D iii

Explanation:

D Isothermal process $\to$ Constant temperature
Adiabatic process $\to$ No Heat exchange
Isochoric process $\to$ Constant volume
Isobaric process $\to$ Constant pressure

AP EAMCET-03.09.2021

Thermodynamics

148428 Assume that you have an ideal gas for which $γ$ $= 1.50$ , initially at $1.0 atm$ pressure When the gas is compressed to half its original volume. then the final pressure. if the compression is isothermals. is

1

4.0 atm

2

2.8 atm

3

2.0 atm

4

6.0 atm

Explanation:

C Given,
Ideal gas, $γ = 1.50$
Initial pressure, $P_{1} = 1 atm$
Initial volume, $V_{1} = V$
Final volume, $V_{2} = \frac{V_{1}}{2} = \frac{V}{2}$ $∵$ Isothermal compression,
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V = P_{2} \times \frac{V}{2}$
$P_{2} = 2 atm$

AP EAMCET-03.09.2021

Thermodynamics

148422 When $50 l$ of air at STP (Standard Temperature and Pressure) is compressed isothermally to $10 l$ , the amount and direction of heat exchange of air is
$[Use P_{atm} = 100 kPa and \ln (0.2) = - 1.61]$

1

8.05 kJ

flows out

2

8.05 kJ

flows in

3

5 kJ

flows out

4

5 kJ

flows in

Explanation:

A Given, Initial volume of air at $STP = 50 l = 50 \times 10^{- 3} m^{3}$ Final Volume of air $= 10 l = 10 \times 10^{- 3} m^{3}$
$P_{atm} = 100 kPa$
Now, Process is Isothermally compressed
$∴$ From $1^{st}$ law of thermodynamics
$Δ Q = Δ U + Δ W$
$∴ Δ Q = Δ W$
Now, for Isothermal process
$Δ W = P_{i} V_{i} \ln (\frac{V_{f}}{V_{i}})$
$= 100 \times 50 \times 10^{- 3} \times \ln (\frac{10 \times 10^{- 3}}{50 \times 10^{- 3}})$
$= 5 \times \ln (0.2)$
$= - 5 \times 1.61$
$Δ Q = - 8.05 kJ$
Negative sign shows that heat rejected from the system
$∴ Δ Q = 8.05 kJ flows out$

TS EAMCET 02.05.2018

Thermodynamics

148424 A graph of pressure (P) against volume (V) of an ideal gas undergoing an isothermal process is:

1 a straight line passing through the origin

2 a parabola

3 a rectangular hyperbola

4 a straight line parallel to pressure axis

Explanation:

C We know that

Ideal gas equation
$PV = nRT$ .....(i)
For isothermal process
$PV =$ Constant
Now, Differentiating both side
$PdV + VdP = 0$
$\frac{dP}{dV} = - \frac{V}{P}$
Note: differential equation of rectangular hyperbola.

AP EAMCET-23.09.2020

Thermodynamics

148425 An ideal gas expands isothermally from volume $V_{1}$ to volume $V_{2}$ . It is then compressed to the original volume $V_{1}$ adiabatically. If $p_{1}, p_{2}$ and $W$ represent the initial pressure, final pressure and the net work done by the gas respectively during the entire process, then

1

p_{1} > p_{2}, W = 0

2

p_{1} > p_{2}, W > 0

3

p_{2} > p_{1}, W > 0

4

p_{2} > p_{1}, W < 0

Explanation:

D Given,

Isothermal
Adiabatic
Expansion
$T_{A} = T_{B}$
Now,
As shown in the above figure
Process $AB =$ Isothermal Expansion
Process $BC =$ Adiabatic Compression From figure it can be said that, area under the curve for isothermal expansion is lesser than area under the curve for adiabatic compression.
$∴$ Work done is negative $(W < 0)$ . Also pressure $(P_{2} > P_{1})$ .

AP EAMCET - 2010

Thermodynamics

1 A iv , B ii , C i , D iii

2 A iv , B i , C ii , D iii

3 A ii , B iii , C iv , D i

4 A ii , B i , C iv , D iii

Explanation:

D Isothermal process $\to$ Constant temperature
Adiabatic process $\to$ No Heat exchange
Isochoric process $\to$ Constant volume
Isobaric process $\to$ Constant pressure

AP EAMCET-03.09.2021

Thermodynamics

148428 Assume that you have an ideal gas for which $γ$ $= 1.50$ , initially at $1.0 atm$ pressure When the gas is compressed to half its original volume. then the final pressure. if the compression is isothermals. is

1

4.0 atm

2

2.8 atm

3

2.0 atm

4

6.0 atm

Explanation:

C Given,
Ideal gas, $γ = 1.50$
Initial pressure, $P_{1} = 1 atm$
Initial volume, $V_{1} = V$
Final volume, $V_{2} = \frac{V_{1}}{2} = \frac{V}{2}$ $∵$ Isothermal compression,
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V = P_{2} \times \frac{V}{2}$
$P_{2} = 2 atm$

AP EAMCET-03.09.2021

Thermodynamics

148422 When $50 l$ of air at STP (Standard Temperature and Pressure) is compressed isothermally to $10 l$ , the amount and direction of heat exchange of air is
$[Use P_{atm} = 100 kPa and \ln (0.2) = - 1.61]$

1

8.05 kJ

flows out

2

8.05 kJ

flows in

3

5 kJ

flows out

4

5 kJ

flows in

Explanation:

A Given, Initial volume of air at $STP = 50 l = 50 \times 10^{- 3} m^{3}$ Final Volume of air $= 10 l = 10 \times 10^{- 3} m^{3}$
$P_{atm} = 100 kPa$
Now, Process is Isothermally compressed
$∴$ From $1^{st}$ law of thermodynamics
$Δ Q = Δ U + Δ W$
$∴ Δ Q = Δ W$
Now, for Isothermal process
$Δ W = P_{i} V_{i} \ln (\frac{V_{f}}{V_{i}})$
$= 100 \times 50 \times 10^{- 3} \times \ln (\frac{10 \times 10^{- 3}}{50 \times 10^{- 3}})$
$= 5 \times \ln (0.2)$
$= - 5 \times 1.61$
$Δ Q = - 8.05 kJ$
Negative sign shows that heat rejected from the system
$∴ Δ Q = 8.05 kJ flows out$

TS EAMCET 02.05.2018

Thermodynamics

148424 A graph of pressure (P) against volume (V) of an ideal gas undergoing an isothermal process is:

1 a straight line passing through the origin

2 a parabola

3 a rectangular hyperbola

4 a straight line parallel to pressure axis

Explanation:

C We know that

Ideal gas equation
$PV = nRT$ .....(i)
For isothermal process
$PV =$ Constant
Now, Differentiating both side
$PdV + VdP = 0$
$\frac{dP}{dV} = - \frac{V}{P}$
Note: differential equation of rectangular hyperbola.

AP EAMCET-23.09.2020

Thermodynamics

148425 An ideal gas expands isothermally from volume $V_{1}$ to volume $V_{2}$ . It is then compressed to the original volume $V_{1}$ adiabatically. If $p_{1}, p_{2}$ and $W$ represent the initial pressure, final pressure and the net work done by the gas respectively during the entire process, then

1

p_{1} > p_{2}, W = 0

2

p_{1} > p_{2}, W > 0

3

p_{2} > p_{1}, W > 0

4

p_{2} > p_{1}, W < 0

Explanation:

D Given,

Isothermal
Adiabatic
Expansion
$T_{A} = T_{B}$
Now,
As shown in the above figure
Process $AB =$ Isothermal Expansion
Process $BC =$ Adiabatic Compression From figure it can be said that, area under the curve for isothermal expansion is lesser than area under the curve for adiabatic compression.
$∴$ Work done is negative $(W < 0)$ . Also pressure $(P_{2} > P_{1})$ .

AP EAMCET - 2010

Thermodynamics

1 A iv , B ii , C i , D iii

2 A iv , B i , C ii , D iii

3 A ii , B iii , C iv , D i

4 A ii , B i , C iv , D iii

Explanation:

D Isothermal process $\to$ Constant temperature
Adiabatic process $\to$ No Heat exchange
Isochoric process $\to$ Constant volume
Isobaric process $\to$ Constant pressure

AP EAMCET-03.09.2021

Thermodynamics

148428 Assume that you have an ideal gas for which $γ$ $= 1.50$ , initially at $1.0 atm$ pressure When the gas is compressed to half its original volume. then the final pressure. if the compression is isothermals. is

1

4.0 atm

2

2.8 atm

3

2.0 atm

4

6.0 atm

Explanation:

C Given,
Ideal gas, $γ = 1.50$
Initial pressure, $P_{1} = 1 atm$
Initial volume, $V_{1} = V$
Final volume, $V_{2} = \frac{V_{1}}{2} = \frac{V}{2}$ $∵$ Isothermal compression,
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V = P_{2} \times \frac{V}{2}$
$P_{2} = 2 atm$

AP EAMCET-03.09.2021

Thermodynamics

148422 When $50 l$ of air at STP (Standard Temperature and Pressure) is compressed isothermally to $10 l$ , the amount and direction of heat exchange of air is
$[Use P_{atm} = 100 kPa and \ln (0.2) = - 1.61]$

1

8.05 kJ

flows out

2

8.05 kJ

flows in

3

5 kJ

flows out

4

5 kJ

flows in

Explanation:

A Given, Initial volume of air at $STP = 50 l = 50 \times 10^{- 3} m^{3}$ Final Volume of air $= 10 l = 10 \times 10^{- 3} m^{3}$
$P_{atm} = 100 kPa$
Now, Process is Isothermally compressed
$∴$ From $1^{st}$ law of thermodynamics
$Δ Q = Δ U + Δ W$
$∴ Δ Q = Δ W$
Now, for Isothermal process
$Δ W = P_{i} V_{i} \ln (\frac{V_{f}}{V_{i}})$
$= 100 \times 50 \times 10^{- 3} \times \ln (\frac{10 \times 10^{- 3}}{50 \times 10^{- 3}})$
$= 5 \times \ln (0.2)$
$= - 5 \times 1.61$
$Δ Q = - 8.05 kJ$
Negative sign shows that heat rejected from the system
$∴ Δ Q = 8.05 kJ flows out$

TS EAMCET 02.05.2018

Thermodynamics

148424 A graph of pressure (P) against volume (V) of an ideal gas undergoing an isothermal process is:

1 a straight line passing through the origin

2 a parabola

3 a rectangular hyperbola

4 a straight line parallel to pressure axis

Explanation:

C We know that

Ideal gas equation
$PV = nRT$ .....(i)
For isothermal process
$PV =$ Constant
Now, Differentiating both side
$PdV + VdP = 0$
$\frac{dP}{dV} = - \frac{V}{P}$
Note: differential equation of rectangular hyperbola.

AP EAMCET-23.09.2020

Thermodynamics

148425 An ideal gas expands isothermally from volume $V_{1}$ to volume $V_{2}$ . It is then compressed to the original volume $V_{1}$ adiabatically. If $p_{1}, p_{2}$ and $W$ represent the initial pressure, final pressure and the net work done by the gas respectively during the entire process, then

1

p_{1} > p_{2}, W = 0

2

p_{1} > p_{2}, W > 0

3

p_{2} > p_{1}, W > 0

4

p_{2} > p_{1}, W < 0

Explanation:

D Given,

Isothermal
Adiabatic
Expansion
$T_{A} = T_{B}$
Now,
As shown in the above figure
Process $AB =$ Isothermal Expansion
Process $BC =$ Adiabatic Compression From figure it can be said that, area under the curve for isothermal expansion is lesser than area under the curve for adiabatic compression.
$∴$ Work done is negative $(W < 0)$ . Also pressure $(P_{2} > P_{1})$ .

AP EAMCET - 2010

Thermodynamics

1 A iv , B ii , C i , D iii

2 A iv , B i , C ii , D iii

3 A ii , B iii , C iv , D i

4 A ii , B i , C iv , D iii

Explanation:

D Isothermal process $\to$ Constant temperature
Adiabatic process $\to$ No Heat exchange
Isochoric process $\to$ Constant volume
Isobaric process $\to$ Constant pressure

AP EAMCET-03.09.2021

Thermodynamics

148428 Assume that you have an ideal gas for which $γ$ $= 1.50$ , initially at $1.0 atm$ pressure When the gas is compressed to half its original volume. then the final pressure. if the compression is isothermals. is

1

4.0 atm

2

2.8 atm

3

2.0 atm

4

6.0 atm

Explanation:

C Given,
Ideal gas, $γ = 1.50$
Initial pressure, $P_{1} = 1 atm$
Initial volume, $V_{1} = V$
Final volume, $V_{2} = \frac{V_{1}}{2} = \frac{V}{2}$ $∵$ Isothermal compression,
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V = P_{2} \times \frac{V}{2}$
$P_{2} = 2 atm$

AP EAMCET-03.09.2021

Thermodynamics

148422 When $50 l$ of air at STP (Standard Temperature and Pressure) is compressed isothermally to $10 l$ , the amount and direction of heat exchange of air is
$[Use P_{atm} = 100 kPa and \ln (0.2) = - 1.61]$

1

8.05 kJ

flows out

2

8.05 kJ

flows in

3

5 kJ

flows out

4

5 kJ

flows in

Explanation:

A Given, Initial volume of air at $STP = 50 l = 50 \times 10^{- 3} m^{3}$ Final Volume of air $= 10 l = 10 \times 10^{- 3} m^{3}$
$P_{atm} = 100 kPa$
Now, Process is Isothermally compressed
$∴$ From $1^{st}$ law of thermodynamics
$Δ Q = Δ U + Δ W$
$∴ Δ Q = Δ W$
Now, for Isothermal process
$Δ W = P_{i} V_{i} \ln (\frac{V_{f}}{V_{i}})$
$= 100 \times 50 \times 10^{- 3} \times \ln (\frac{10 \times 10^{- 3}}{50 \times 10^{- 3}})$
$= 5 \times \ln (0.2)$
$= - 5 \times 1.61$
$Δ Q = - 8.05 kJ$
Negative sign shows that heat rejected from the system
$∴ Δ Q = 8.05 kJ flows out$

TS EAMCET 02.05.2018

Thermodynamics

148424 A graph of pressure (P) against volume (V) of an ideal gas undergoing an isothermal process is:

1 a straight line passing through the origin

2 a parabola

3 a rectangular hyperbola

4 a straight line parallel to pressure axis

Explanation:

C We know that

Ideal gas equation
$PV = nRT$ .....(i)
For isothermal process
$PV =$ Constant
Now, Differentiating both side
$PdV + VdP = 0$
$\frac{dP}{dV} = - \frac{V}{P}$
Note: differential equation of rectangular hyperbola.

AP EAMCET-23.09.2020

Thermodynamics

148425 An ideal gas expands isothermally from volume $V_{1}$ to volume $V_{2}$ . It is then compressed to the original volume $V_{1}$ adiabatically. If $p_{1}, p_{2}$ and $W$ represent the initial pressure, final pressure and the net work done by the gas respectively during the entire process, then

1

p_{1} > p_{2}, W = 0

2

p_{1} > p_{2}, W > 0

3

p_{2} > p_{1}, W > 0

4

p_{2} > p_{1}, W < 0

Explanation:

D Given,

Isothermal
Adiabatic
Expansion
$T_{A} = T_{B}$
Now,
As shown in the above figure
Process $AB =$ Isothermal Expansion
Process $BC =$ Adiabatic Compression From figure it can be said that, area under the curve for isothermal expansion is lesser than area under the curve for adiabatic compression.
$∴$ Work done is negative $(W < 0)$ . Also pressure $(P_{2} > P_{1})$ .

AP EAMCET - 2010

Thermodynamics

1 A iv , B ii , C i , D iii

2 A iv , B i , C ii , D iii

3 A ii , B iii , C iv , D i

4 A ii , B i , C iv , D iii

Explanation:

D Isothermal process $\to$ Constant temperature
Adiabatic process $\to$ No Heat exchange
Isochoric process $\to$ Constant volume
Isobaric process $\to$ Constant pressure

AP EAMCET-03.09.2021

Thermodynamics

148428 Assume that you have an ideal gas for which $γ$ $= 1.50$ , initially at $1.0 atm$ pressure When the gas is compressed to half its original volume. then the final pressure. if the compression is isothermals. is

1

4.0 atm

2

2.8 atm

3

2.0 atm

4

6.0 atm

Explanation:

C Given,
Ideal gas, $γ = 1.50$
Initial pressure, $P_{1} = 1 atm$
Initial volume, $V_{1} = V$
Final volume, $V_{2} = \frac{V_{1}}{2} = \frac{V}{2}$ $∵$ Isothermal compression,
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V = P_{2} \times \frac{V}{2}$
$P_{2} = 2 atm$

AP EAMCET-03.09.2021

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