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, $\mathrm{V}_{1}=\mathrm{V}, \quad \mathrm{V}_{2}=2 \mathrm{~V}$ For isothermal $\Delta \mathrm{U}=0$ $\mathrm{dW}=\mathrm{P} \Delta \mathrm{V}$ $\mathrm{dW}=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ $\mathrm{dW}=\mathrm{P}(2 \mathrm{~V}-\mathrm{V})$ $\mathrm{dW}=\mathrm{PV}$ Hence, the work done by the gas is positive.
JIPMER-2016
Thermodynamics
148419
The slope of isothermal and adiabatic curves are related as
C In the slope of isothermal curve, $\mathrm{T}=$ constant $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=$ constant $\frac{\mathrm{PdV}}{\mathrm{dV}}+\frac{\mathrm{VdP}}{\mathrm{dV}}=0$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\left(\frac{-\mathrm{P}}{\mathrm{V}}\right)$ Slope of isothermal curve on $\mathrm{P}-\mathrm{V}$ diagram $=(-\mathrm{P} / \mathrm{V})$ Slope of adiabatic curve on P-V diagram $\mathrm{PV}^{\gamma}=\mathrm{C}$ Differentiating with respect to $\mathrm{V}$, $\mathrm{P} \frac{\mathrm{d}\left(\mathrm{V}^{\gamma}\right)}{\mathrm{dV}}+\mathrm{V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=0$ $\mathrm{~V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \frac{\mathrm{d}\left(\mathrm{V}^{\gamma}\right)}{\mathrm{dV}}$ $\mathrm{V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \gamma \mathrm{V}^{\gamma-1}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\mathrm{P} \gamma \mathrm{V}^{\gamma-1}}{\mathrm{~V}^{\gamma}}=-\mathrm{P} \gamma \mathrm{V}^{\gamma-1-\gamma}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \gamma \mathrm{V}^{-1}$ $\therefore \quad \frac{\mathrm{dP}}{\mathrm{dV}}=-\gamma \frac{\mathrm{P}}{\mathrm{V}}$ Hence, adiabatic slope $=\gamma$ 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 $\mathrm{m}_{1} \lt \mathrm{m}_{2}$
3 $\mathrm{m}_{1}=\mathrm{m}_{2}$
4 $\mathrm{m}_{1}>\mathrm{m}_{2}$
Explanation:
B From the ideal gas equation for the slope $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$ For slope (I) :- $\mathrm{P}=\frac{\mathrm{m}_{1}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{1}}$ For slope (II):- $\mathrm{P}=\frac{\mathrm{m}_{2}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{2}}$ From equation (i) and (ii), $\frac{\mathrm{m}_{1}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{MV}_{2}} \mathrm{RT}$ $\frac{\mathrm{m}_{1}}{\mathrm{~V}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{~V}_{2}}$ $\mathrm{~V} \propto \mathrm{m}$ From the figure $\mathrm{V}_{2}>\mathrm{V}_{1}$ $\therefore \quad \mathrm{m}_{2}>\mathrm{m}_{1}$
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 $\mathrm{PV}=$ constant Differentiating both side $\mathrm{PdV}+\mathrm{VdP}=0$ $\mathrm{VdP}=-\mathrm{PdV}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\mathrm{P}}{\mathrm{V}}$ For adiabatic Process $\mathrm{PV}^{\gamma}=\text { Constant }$ Differentiating both side $\mathrm{P}_{\gamma} \mathrm{V}^{\gamma-1} \mathrm{dV}+\mathrm{V}^{\gamma} \mathrm{dP}=0$ $\mathrm{~V}^{\gamma} \mathrm{dP}=-\gamma \mathrm{PV}^{\gamma-1} \mathrm{dV}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\gamma \mathrm{PV}^{\gamma-1}}{\mathrm{~V}^{\gamma}}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\gamma \mathrm{P}}{\mathrm{V}}$ $\therefore \quad \left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)_{\text {isothermal }} \lt \left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)_{\text {Adiabatic }}$ $\therefore$ Slope of an Isothermal curve is always less than adiabatic curve.
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, $\mathrm{V}_{1}=\mathrm{V}, \quad \mathrm{V}_{2}=2 \mathrm{~V}$ For isothermal $\Delta \mathrm{U}=0$ $\mathrm{dW}=\mathrm{P} \Delta \mathrm{V}$ $\mathrm{dW}=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ $\mathrm{dW}=\mathrm{P}(2 \mathrm{~V}-\mathrm{V})$ $\mathrm{dW}=\mathrm{PV}$ Hence, the work done by the gas is positive.
JIPMER-2016
Thermodynamics
148419
The slope of isothermal and adiabatic curves are related as
C In the slope of isothermal curve, $\mathrm{T}=$ constant $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=$ constant $\frac{\mathrm{PdV}}{\mathrm{dV}}+\frac{\mathrm{VdP}}{\mathrm{dV}}=0$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\left(\frac{-\mathrm{P}}{\mathrm{V}}\right)$ Slope of isothermal curve on $\mathrm{P}-\mathrm{V}$ diagram $=(-\mathrm{P} / \mathrm{V})$ Slope of adiabatic curve on P-V diagram $\mathrm{PV}^{\gamma}=\mathrm{C}$ Differentiating with respect to $\mathrm{V}$, $\mathrm{P} \frac{\mathrm{d}\left(\mathrm{V}^{\gamma}\right)}{\mathrm{dV}}+\mathrm{V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=0$ $\mathrm{~V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \frac{\mathrm{d}\left(\mathrm{V}^{\gamma}\right)}{\mathrm{dV}}$ $\mathrm{V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \gamma \mathrm{V}^{\gamma-1}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\mathrm{P} \gamma \mathrm{V}^{\gamma-1}}{\mathrm{~V}^{\gamma}}=-\mathrm{P} \gamma \mathrm{V}^{\gamma-1-\gamma}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \gamma \mathrm{V}^{-1}$ $\therefore \quad \frac{\mathrm{dP}}{\mathrm{dV}}=-\gamma \frac{\mathrm{P}}{\mathrm{V}}$ Hence, adiabatic slope $=\gamma$ 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 $\mathrm{m}_{1} \lt \mathrm{m}_{2}$
3 $\mathrm{m}_{1}=\mathrm{m}_{2}$
4 $\mathrm{m}_{1}>\mathrm{m}_{2}$
Explanation:
B From the ideal gas equation for the slope $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$ For slope (I) :- $\mathrm{P}=\frac{\mathrm{m}_{1}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{1}}$ For slope (II):- $\mathrm{P}=\frac{\mathrm{m}_{2}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{2}}$ From equation (i) and (ii), $\frac{\mathrm{m}_{1}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{MV}_{2}} \mathrm{RT}$ $\frac{\mathrm{m}_{1}}{\mathrm{~V}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{~V}_{2}}$ $\mathrm{~V} \propto \mathrm{m}$ From the figure $\mathrm{V}_{2}>\mathrm{V}_{1}$ $\therefore \quad \mathrm{m}_{2}>\mathrm{m}_{1}$
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 $\mathrm{PV}=$ constant Differentiating both side $\mathrm{PdV}+\mathrm{VdP}=0$ $\mathrm{VdP}=-\mathrm{PdV}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\mathrm{P}}{\mathrm{V}}$ For adiabatic Process $\mathrm{PV}^{\gamma}=\text { Constant }$ Differentiating both side $\mathrm{P}_{\gamma} \mathrm{V}^{\gamma-1} \mathrm{dV}+\mathrm{V}^{\gamma} \mathrm{dP}=0$ $\mathrm{~V}^{\gamma} \mathrm{dP}=-\gamma \mathrm{PV}^{\gamma-1} \mathrm{dV}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\gamma \mathrm{PV}^{\gamma-1}}{\mathrm{~V}^{\gamma}}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\gamma \mathrm{P}}{\mathrm{V}}$ $\therefore \quad \left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)_{\text {isothermal }} \lt \left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)_{\text {Adiabatic }}$ $\therefore$ Slope of an Isothermal curve is always less than adiabatic curve.
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, $\mathrm{V}_{1}=\mathrm{V}, \quad \mathrm{V}_{2}=2 \mathrm{~V}$ For isothermal $\Delta \mathrm{U}=0$ $\mathrm{dW}=\mathrm{P} \Delta \mathrm{V}$ $\mathrm{dW}=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ $\mathrm{dW}=\mathrm{P}(2 \mathrm{~V}-\mathrm{V})$ $\mathrm{dW}=\mathrm{PV}$ Hence, the work done by the gas is positive.
JIPMER-2016
Thermodynamics
148419
The slope of isothermal and adiabatic curves are related as
C In the slope of isothermal curve, $\mathrm{T}=$ constant $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=$ constant $\frac{\mathrm{PdV}}{\mathrm{dV}}+\frac{\mathrm{VdP}}{\mathrm{dV}}=0$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\left(\frac{-\mathrm{P}}{\mathrm{V}}\right)$ Slope of isothermal curve on $\mathrm{P}-\mathrm{V}$ diagram $=(-\mathrm{P} / \mathrm{V})$ Slope of adiabatic curve on P-V diagram $\mathrm{PV}^{\gamma}=\mathrm{C}$ Differentiating with respect to $\mathrm{V}$, $\mathrm{P} \frac{\mathrm{d}\left(\mathrm{V}^{\gamma}\right)}{\mathrm{dV}}+\mathrm{V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=0$ $\mathrm{~V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \frac{\mathrm{d}\left(\mathrm{V}^{\gamma}\right)}{\mathrm{dV}}$ $\mathrm{V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \gamma \mathrm{V}^{\gamma-1}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\mathrm{P} \gamma \mathrm{V}^{\gamma-1}}{\mathrm{~V}^{\gamma}}=-\mathrm{P} \gamma \mathrm{V}^{\gamma-1-\gamma}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \gamma \mathrm{V}^{-1}$ $\therefore \quad \frac{\mathrm{dP}}{\mathrm{dV}}=-\gamma \frac{\mathrm{P}}{\mathrm{V}}$ Hence, adiabatic slope $=\gamma$ 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 $\mathrm{m}_{1} \lt \mathrm{m}_{2}$
3 $\mathrm{m}_{1}=\mathrm{m}_{2}$
4 $\mathrm{m}_{1}>\mathrm{m}_{2}$
Explanation:
B From the ideal gas equation for the slope $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$ For slope (I) :- $\mathrm{P}=\frac{\mathrm{m}_{1}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{1}}$ For slope (II):- $\mathrm{P}=\frac{\mathrm{m}_{2}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{2}}$ From equation (i) and (ii), $\frac{\mathrm{m}_{1}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{MV}_{2}} \mathrm{RT}$ $\frac{\mathrm{m}_{1}}{\mathrm{~V}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{~V}_{2}}$ $\mathrm{~V} \propto \mathrm{m}$ From the figure $\mathrm{V}_{2}>\mathrm{V}_{1}$ $\therefore \quad \mathrm{m}_{2}>\mathrm{m}_{1}$
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 $\mathrm{PV}=$ constant Differentiating both side $\mathrm{PdV}+\mathrm{VdP}=0$ $\mathrm{VdP}=-\mathrm{PdV}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\mathrm{P}}{\mathrm{V}}$ For adiabatic Process $\mathrm{PV}^{\gamma}=\text { Constant }$ Differentiating both side $\mathrm{P}_{\gamma} \mathrm{V}^{\gamma-1} \mathrm{dV}+\mathrm{V}^{\gamma} \mathrm{dP}=0$ $\mathrm{~V}^{\gamma} \mathrm{dP}=-\gamma \mathrm{PV}^{\gamma-1} \mathrm{dV}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\gamma \mathrm{PV}^{\gamma-1}}{\mathrm{~V}^{\gamma}}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\gamma \mathrm{P}}{\mathrm{V}}$ $\therefore \quad \left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)_{\text {isothermal }} \lt \left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)_{\text {Adiabatic }}$ $\therefore$ Slope of an Isothermal curve is always less than adiabatic curve.
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, $\mathrm{V}_{1}=\mathrm{V}, \quad \mathrm{V}_{2}=2 \mathrm{~V}$ For isothermal $\Delta \mathrm{U}=0$ $\mathrm{dW}=\mathrm{P} \Delta \mathrm{V}$ $\mathrm{dW}=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ $\mathrm{dW}=\mathrm{P}(2 \mathrm{~V}-\mathrm{V})$ $\mathrm{dW}=\mathrm{PV}$ Hence, the work done by the gas is positive.
JIPMER-2016
Thermodynamics
148419
The slope of isothermal and adiabatic curves are related as
C In the slope of isothermal curve, $\mathrm{T}=$ constant $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=$ constant $\frac{\mathrm{PdV}}{\mathrm{dV}}+\frac{\mathrm{VdP}}{\mathrm{dV}}=0$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\left(\frac{-\mathrm{P}}{\mathrm{V}}\right)$ Slope of isothermal curve on $\mathrm{P}-\mathrm{V}$ diagram $=(-\mathrm{P} / \mathrm{V})$ Slope of adiabatic curve on P-V diagram $\mathrm{PV}^{\gamma}=\mathrm{C}$ Differentiating with respect to $\mathrm{V}$, $\mathrm{P} \frac{\mathrm{d}\left(\mathrm{V}^{\gamma}\right)}{\mathrm{dV}}+\mathrm{V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=0$ $\mathrm{~V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \frac{\mathrm{d}\left(\mathrm{V}^{\gamma}\right)}{\mathrm{dV}}$ $\mathrm{V}^{\gamma} \frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \gamma \mathrm{V}^{\gamma-1}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\mathrm{P} \gamma \mathrm{V}^{\gamma-1}}{\mathrm{~V}^{\gamma}}=-\mathrm{P} \gamma \mathrm{V}^{\gamma-1-\gamma}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=-\mathrm{P} \gamma \mathrm{V}^{-1}$ $\therefore \quad \frac{\mathrm{dP}}{\mathrm{dV}}=-\gamma \frac{\mathrm{P}}{\mathrm{V}}$ Hence, adiabatic slope $=\gamma$ 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 $\mathrm{m}_{1} \lt \mathrm{m}_{2}$
3 $\mathrm{m}_{1}=\mathrm{m}_{2}$
4 $\mathrm{m}_{1}>\mathrm{m}_{2}$
Explanation:
B From the ideal gas equation for the slope $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$ For slope (I) :- $\mathrm{P}=\frac{\mathrm{m}_{1}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{1}}$ For slope (II):- $\mathrm{P}=\frac{\mathrm{m}_{2}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{2}}$ From equation (i) and (ii), $\frac{\mathrm{m}_{1}}{\mathrm{M}} \frac{\mathrm{RT}}{\mathrm{V}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{MV}_{2}} \mathrm{RT}$ $\frac{\mathrm{m}_{1}}{\mathrm{~V}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{~V}_{2}}$ $\mathrm{~V} \propto \mathrm{m}$ From the figure $\mathrm{V}_{2}>\mathrm{V}_{1}$ $\therefore \quad \mathrm{m}_{2}>\mathrm{m}_{1}$
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 $\mathrm{PV}=$ constant Differentiating both side $\mathrm{PdV}+\mathrm{VdP}=0$ $\mathrm{VdP}=-\mathrm{PdV}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\mathrm{P}}{\mathrm{V}}$ For adiabatic Process $\mathrm{PV}^{\gamma}=\text { Constant }$ Differentiating both side $\mathrm{P}_{\gamma} \mathrm{V}^{\gamma-1} \mathrm{dV}+\mathrm{V}^{\gamma} \mathrm{dP}=0$ $\mathrm{~V}^{\gamma} \mathrm{dP}=-\gamma \mathrm{PV}^{\gamma-1} \mathrm{dV}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\gamma \mathrm{PV}^{\gamma-1}}{\mathrm{~V}^{\gamma}}$ $\frac{\mathrm{dP}}{\mathrm{dV}}=\frac{-\gamma \mathrm{P}}{\mathrm{V}}$ $\therefore \quad \left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)_{\text {isothermal }} \lt \left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)_{\text {Adiabatic }}$ $\therefore$ Slope of an Isothermal curve is always less than adiabatic curve.
