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Thermodynamics

148418 The volume of an ideal gas is doubled in an isothermal process. Then, which of the following is true ?

1 Work done by the gas is positive

2 Work done by the gas is negative

3 Internal energy of the system decreases

4 Internal energy of the system increases

Explanation:

A Given that,
$V_{1} = V, V_{2} = 2 V$
For isothermal $Δ U = 0$
$dW = P Δ V$
$dW = P (V_{2} - V_{1})$
$dW = P (2 V - V)$
$dW = PV$
Hence, the work done by the gas is positive.

JIPMER-2016

Thermodynamics

148419 The slope of isothermal and adiabatic curves are related as

1 Isothermal curve slope

=

adiabatic curve slope

2 Isothermal curve slope

= γ \times

adiabatic curve slope

3 Adiabatic curve slope

= γ \times

Isothermal curve slope

4 Adiabatic curve slope

= \frac{1}{2} \times

Isothermal curve slope

Explanation:

C In the slope of isothermal curve,
$T =$ constant
$PV = nRT$
$PV =$ constant
$\frac{PdV}{dV} + \frac{VdP}{dV} = 0$
$\frac{dP}{dV} = (\frac{- P}{V})$
Slope of isothermal curve on $P - V$ diagram $= (- P / V)$
Slope of adiabatic curve on P-V diagram
${PV}^{γ} = C$
Differentiating with respect to $V$ ,
$P \frac{d (V^{γ})}{dV} + V^{γ} \frac{dP}{dV} = 0$
$V^{γ} \frac{dP}{dV} = - P \frac{d (V^{γ})}{dV}$
$V^{γ} \frac{dP}{dV} = - P γ V^{γ - 1}$
$\frac{dP}{dV} = \frac{- P γ V^{γ - 1}}{V^{γ}} = - P γ V^{γ - 1 - γ}$
$\frac{dP}{dV} = - P γ V^{- 1}$
$∴ \frac{dP}{dV} = - γ \frac{P}{V}$
Hence, adiabatic slope $= γ$ times the slope of isothermal curve.

UP CPMT 1971

Thermodynamics

148420 Two different isotherms representing the relationship between pressure $p$ and volume $V$ at a given temperature of the same ideal gas are shown for masses $m_{1}$ and $m_{2}$ , then

1 Nothing can be predicted

2

m_{1} < m_{2}

3

m_{1} = m_{2}

4

m_{1} > m_{2}

Explanation:

B
From the ideal gas equation for the slope $m_{1}$ and $m_{2}$
$PV = nRT$
$PV = \frac{m}{M} RT$
For slope (I) :-
$P = \frac{m_{1}}{M} \frac{RT}{V_{1}}$
For slope (II):-
$P = \frac{m_{2}}{M} \frac{RT}{V_{2}}$
From equation (i) and (ii),
$\frac{m_{1}}{M} \frac{RT}{V_{1}} = \frac{m_{2}}{{MV}_{2}} RT$
$\frac{m_{1}}{V_{1}} = \frac{m_{2}}{V_{2}}$
$V \propto m$
From the figure $V_{2} > V_{1}$
$∴ m_{2} > m_{1}$

JIPMER-2016

Thermodynamics

148421 Slope of an isothermal curve is always

1 equal to adiabatic curve

2 greater than adiabatic curve

3 less than adiabatic curve

4 cannot be determined

Explanation:

C We know that, for Isothermal process $PV =$ constant
Differentiating both side
$PdV + VdP = 0$
$VdP = - PdV$
$\frac{dP}{dV} = \frac{- P}{V}$ For adiabatic Process
${PV}^{γ} = Constant$
Differentiating both side
$P_{γ} V^{γ - 1} dV + V^{γ} dP = 0$
$V^{γ} dP = - γ {PV}^{γ - 1} dV$
$\frac{dP}{dV} = \frac{- γ {PV}^{γ - 1}}{V^{γ}}$
$\frac{dP}{dV} = \frac{- γ P}{V}$
$∴ {(\frac{dP}{dV})}_{isothermal} < {(\frac{dP}{dV})}_{Adiabatic}$
$∴$ Slope of an Isothermal curve is always less than adiabatic curve.

WB JEE-2007

Thermodynamics

148418 The volume of an ideal gas is doubled in an isothermal process. Then, which of the following is true ?

1 Work done by the gas is positive

2 Work done by the gas is negative

3 Internal energy of the system decreases

4 Internal energy of the system increases

Explanation:

A Given that,
$V_{1} = V, V_{2} = 2 V$
For isothermal $Δ U = 0$
$dW = P Δ V$
$dW = P (V_{2} - V_{1})$
$dW = P (2 V - V)$
$dW = PV$
Hence, the work done by the gas is positive.

JIPMER-2016

Thermodynamics

148419 The slope of isothermal and adiabatic curves are related as

1 Isothermal curve slope

=

adiabatic curve slope

2 Isothermal curve slope

= γ \times

adiabatic curve slope

3 Adiabatic curve slope

= γ \times

Isothermal curve slope

4 Adiabatic curve slope

= \frac{1}{2} \times

Isothermal curve slope

Explanation:

C In the slope of isothermal curve,
$T =$ constant
$PV = nRT$
$PV =$ constant
$\frac{PdV}{dV} + \frac{VdP}{dV} = 0$
$\frac{dP}{dV} = (\frac{- P}{V})$
Slope of isothermal curve on $P - V$ diagram $= (- P / V)$
Slope of adiabatic curve on P-V diagram
${PV}^{γ} = C$
Differentiating with respect to $V$ ,
$P \frac{d (V^{γ})}{dV} + V^{γ} \frac{dP}{dV} = 0$
$V^{γ} \frac{dP}{dV} = - P \frac{d (V^{γ})}{dV}$
$V^{γ} \frac{dP}{dV} = - P γ V^{γ - 1}$
$\frac{dP}{dV} = \frac{- P γ V^{γ - 1}}{V^{γ}} = - P γ V^{γ - 1 - γ}$
$\frac{dP}{dV} = - P γ V^{- 1}$
$∴ \frac{dP}{dV} = - γ \frac{P}{V}$
Hence, adiabatic slope $= γ$ times the slope of isothermal curve.

UP CPMT 1971

Thermodynamics

148420 Two different isotherms representing the relationship between pressure $p$ and volume $V$ at a given temperature of the same ideal gas are shown for masses $m_{1}$ and $m_{2}$ , then

1 Nothing can be predicted

2

m_{1} < m_{2}

3

m_{1} = m_{2}

4

m_{1} > m_{2}

Explanation:

B
From the ideal gas equation for the slope $m_{1}$ and $m_{2}$
$PV = nRT$
$PV = \frac{m}{M} RT$
For slope (I) :-
$P = \frac{m_{1}}{M} \frac{RT}{V_{1}}$
For slope (II):-
$P = \frac{m_{2}}{M} \frac{RT}{V_{2}}$
From equation (i) and (ii),
$\frac{m_{1}}{M} \frac{RT}{V_{1}} = \frac{m_{2}}{{MV}_{2}} RT$
$\frac{m_{1}}{V_{1}} = \frac{m_{2}}{V_{2}}$
$V \propto m$
From the figure $V_{2} > V_{1}$
$∴ m_{2} > m_{1}$

JIPMER-2016

Thermodynamics

148421 Slope of an isothermal curve is always

1 equal to adiabatic curve

2 greater than adiabatic curve

3 less than adiabatic curve

4 cannot be determined

Explanation:

C We know that, for Isothermal process $PV =$ constant
Differentiating both side
$PdV + VdP = 0$
$VdP = - PdV$
$\frac{dP}{dV} = \frac{- P}{V}$ For adiabatic Process
${PV}^{γ} = Constant$
Differentiating both side
$P_{γ} V^{γ - 1} dV + V^{γ} dP = 0$
$V^{γ} dP = - γ {PV}^{γ - 1} dV$
$\frac{dP}{dV} = \frac{- γ {PV}^{γ - 1}}{V^{γ}}$
$\frac{dP}{dV} = \frac{- γ P}{V}$
$∴ {(\frac{dP}{dV})}_{isothermal} < {(\frac{dP}{dV})}_{Adiabatic}$
$∴$ Slope of an Isothermal curve is always less than adiabatic curve.

WB JEE-2007

Thermodynamics

148418 The volume of an ideal gas is doubled in an isothermal process. Then, which of the following is true ?

1 Work done by the gas is positive

2 Work done by the gas is negative

3 Internal energy of the system decreases

4 Internal energy of the system increases

Explanation:

A Given that,
$V_{1} = V, V_{2} = 2 V$
For isothermal $Δ U = 0$
$dW = P Δ V$
$dW = P (V_{2} - V_{1})$
$dW = P (2 V - V)$
$dW = PV$
Hence, the work done by the gas is positive.

JIPMER-2016

Thermodynamics

148419 The slope of isothermal and adiabatic curves are related as

1 Isothermal curve slope

=

adiabatic curve slope

2 Isothermal curve slope

= γ \times

adiabatic curve slope

3 Adiabatic curve slope

= γ \times

Isothermal curve slope

4 Adiabatic curve slope

= \frac{1}{2} \times

Isothermal curve slope

Explanation:

C In the slope of isothermal curve,
$T =$ constant
$PV = nRT$
$PV =$ constant
$\frac{PdV}{dV} + \frac{VdP}{dV} = 0$
$\frac{dP}{dV} = (\frac{- P}{V})$
Slope of isothermal curve on $P - V$ diagram $= (- P / V)$
Slope of adiabatic curve on P-V diagram
${PV}^{γ} = C$
Differentiating with respect to $V$ ,
$P \frac{d (V^{γ})}{dV} + V^{γ} \frac{dP}{dV} = 0$
$V^{γ} \frac{dP}{dV} = - P \frac{d (V^{γ})}{dV}$
$V^{γ} \frac{dP}{dV} = - P γ V^{γ - 1}$
$\frac{dP}{dV} = \frac{- P γ V^{γ - 1}}{V^{γ}} = - P γ V^{γ - 1 - γ}$
$\frac{dP}{dV} = - P γ V^{- 1}$
$∴ \frac{dP}{dV} = - γ \frac{P}{V}$
Hence, adiabatic slope $= γ$ times the slope of isothermal curve.

UP CPMT 1971

Thermodynamics

148420 Two different isotherms representing the relationship between pressure $p$ and volume $V$ at a given temperature of the same ideal gas are shown for masses $m_{1}$ and $m_{2}$ , then

1 Nothing can be predicted

2

m_{1} < m_{2}

3

m_{1} = m_{2}

4

m_{1} > m_{2}

Explanation:

B
From the ideal gas equation for the slope $m_{1}$ and $m_{2}$
$PV = nRT$
$PV = \frac{m}{M} RT$
For slope (I) :-
$P = \frac{m_{1}}{M} \frac{RT}{V_{1}}$
For slope (II):-
$P = \frac{m_{2}}{M} \frac{RT}{V_{2}}$
From equation (i) and (ii),
$\frac{m_{1}}{M} \frac{RT}{V_{1}} = \frac{m_{2}}{{MV}_{2}} RT$
$\frac{m_{1}}{V_{1}} = \frac{m_{2}}{V_{2}}$
$V \propto m$
From the figure $V_{2} > V_{1}$
$∴ m_{2} > m_{1}$

JIPMER-2016

Thermodynamics

148421 Slope of an isothermal curve is always

1 equal to adiabatic curve

2 greater than adiabatic curve

3 less than adiabatic curve

4 cannot be determined

Explanation:

C We know that, for Isothermal process $PV =$ constant
Differentiating both side
$PdV + VdP = 0$
$VdP = - PdV$
$\frac{dP}{dV} = \frac{- P}{V}$ For adiabatic Process
${PV}^{γ} = Constant$
Differentiating both side
$P_{γ} V^{γ - 1} dV + V^{γ} dP = 0$
$V^{γ} dP = - γ {PV}^{γ - 1} dV$
$\frac{dP}{dV} = \frac{- γ {PV}^{γ - 1}}{V^{γ}}$
$\frac{dP}{dV} = \frac{- γ P}{V}$
$∴ {(\frac{dP}{dV})}_{isothermal} < {(\frac{dP}{dV})}_{Adiabatic}$
$∴$ Slope of an Isothermal curve is always less than adiabatic curve.

WB JEE-2007

Thermodynamics

148418 The volume of an ideal gas is doubled in an isothermal process. Then, which of the following is true ?

1 Work done by the gas is positive

2 Work done by the gas is negative

3 Internal energy of the system decreases

4 Internal energy of the system increases

Explanation:

A Given that,
$V_{1} = V, V_{2} = 2 V$
For isothermal $Δ U = 0$
$dW = P Δ V$
$dW = P (V_{2} - V_{1})$
$dW = P (2 V - V)$
$dW = PV$
Hence, the work done by the gas is positive.

JIPMER-2016

Thermodynamics

148419 The slope of isothermal and adiabatic curves are related as

1 Isothermal curve slope

=

adiabatic curve slope

2 Isothermal curve slope

= γ \times

adiabatic curve slope

3 Adiabatic curve slope

= γ \times

Isothermal curve slope

4 Adiabatic curve slope

= \frac{1}{2} \times

Isothermal curve slope

Explanation:

C In the slope of isothermal curve,
$T =$ constant
$PV = nRT$
$PV =$ constant
$\frac{PdV}{dV} + \frac{VdP}{dV} = 0$
$\frac{dP}{dV} = (\frac{- P}{V})$
Slope of isothermal curve on $P - V$ diagram $= (- P / V)$
Slope of adiabatic curve on P-V diagram
${PV}^{γ} = C$
Differentiating with respect to $V$ ,
$P \frac{d (V^{γ})}{dV} + V^{γ} \frac{dP}{dV} = 0$
$V^{γ} \frac{dP}{dV} = - P \frac{d (V^{γ})}{dV}$
$V^{γ} \frac{dP}{dV} = - P γ V^{γ - 1}$
$\frac{dP}{dV} = \frac{- P γ V^{γ - 1}}{V^{γ}} = - P γ V^{γ - 1 - γ}$
$\frac{dP}{dV} = - P γ V^{- 1}$
$∴ \frac{dP}{dV} = - γ \frac{P}{V}$
Hence, adiabatic slope $= γ$ times the slope of isothermal curve.

UP CPMT 1971

Thermodynamics

148420 Two different isotherms representing the relationship between pressure $p$ and volume $V$ at a given temperature of the same ideal gas are shown for masses $m_{1}$ and $m_{2}$ , then

1 Nothing can be predicted

2

m_{1} < m_{2}

3

m_{1} = m_{2}

4

m_{1} > m_{2}

Explanation:

B
From the ideal gas equation for the slope $m_{1}$ and $m_{2}$
$PV = nRT$
$PV = \frac{m}{M} RT$
For slope (I) :-
$P = \frac{m_{1}}{M} \frac{RT}{V_{1}}$
For slope (II):-
$P = \frac{m_{2}}{M} \frac{RT}{V_{2}}$
From equation (i) and (ii),
$\frac{m_{1}}{M} \frac{RT}{V_{1}} = \frac{m_{2}}{{MV}_{2}} RT$
$\frac{m_{1}}{V_{1}} = \frac{m_{2}}{V_{2}}$
$V \propto m$
From the figure $V_{2} > V_{1}$
$∴ m_{2} > m_{1}$

JIPMER-2016

Thermodynamics

148421 Slope of an isothermal curve is always

1 equal to adiabatic curve

2 greater than adiabatic curve

3 less than adiabatic curve

4 cannot be determined

Explanation:

C We know that, for Isothermal process $PV =$ constant
Differentiating both side
$PdV + VdP = 0$
$VdP = - PdV$
$\frac{dP}{dV} = \frac{- P}{V}$ For adiabatic Process
${PV}^{γ} = Constant$
Differentiating both side
$P_{γ} V^{γ - 1} dV + V^{γ} dP = 0$
$V^{γ} dP = - γ {PV}^{γ - 1} dV$
$\frac{dP}{dV} = \frac{- γ {PV}^{γ - 1}}{V^{γ}}$
$\frac{dP}{dV} = \frac{- γ P}{V}$
$∴ {(\frac{dP}{dV})}_{isothermal} < {(\frac{dP}{dV})}_{Adiabatic}$
$∴$ Slope of an Isothermal curve is always less than adiabatic curve.

WB JEE-2007