148452
In an adiabatic expansion of an ideal gas the product of pressure and volume :
1 decreases
2 increases
3 remains constant
4 at first increases and then decreases
Explanation:
A Ideal gas equation is given by, $\mathrm{PV}=\mathrm{nRT}$ $\because \mathrm{T}$ is decreasing, $\mathrm{PV}$ also decreases. If $\mathrm{T}$ decreases $\mathrm{PV}$ also decreases because in an adiabatic expansion. Internal energy decreases and temperature also decreases.
Karnataka CET-2020
Thermodynamics
148461
In an adiabatic process
1 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
2 $\mathrm{TV}^{\gamma-1}=$ constant
3 $\mathrm{PV}=$ constant
4 All of these
Explanation:
B For an adiabatic process, $\mathrm{PV}^{\gamma}=$ constant. Using the ideal gas equation and substituting $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$ In the above equation gives $\mathrm{TV}^{\gamma-1}=$ constant Where $\gamma$ is the specific heat ratio
UP CPMT-2003
Thermodynamics
148463
During adiabatic expansion, the increase in the volume is associated with
1 increase in pressure and decrease in $\mathrm{T}$
2 decrease in pressure and increase in $\mathrm{T}$
3 increase in pressure and $\mathrm{T}$
4 decrease in pressure and $\mathrm{T}$
Explanation:
D $\mathrm{PV}^{\gamma}=$ constant And, $\mathrm{TV}^{\gamma-1}=$ constant From the above equation increasing in volume pressure and temperature both are decreases.
TS EAMCET 30.07.2022
Thermodynamics
148464
An ideal gas undergoes an adiabatic process. If the pressure of the gas is reduced by $0.1 \%$ then the volume is changed by $\left(\right.$ Given $\left.\gamma=\frac{C_{p}}{C_{v}}=5 / 3\right)$
1 $0.1 \%$
2 $0.05 \%$
3 $0.06 \%$
4 $-0.05 \%$
Explanation:
C From the relation, $\mathrm{PV}^{\gamma}=\mathrm{K}$ Now differentiating we get, $\mathrm{PV}^{\gamma-1} \mathrm{dV}+\mathrm{dP} \cdot \mathrm{V}^{\gamma}=0$ By making proper rearrangements we get, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{\mathrm{dP}}{\gamma \mathrm{P}}$ In adiabatic process on a gas with $\gamma=1.4$, the pressure of the is reduced by $0.1 \%$. The volume (decreases) changes by about, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{0.1}{1.4} \times 100=0.06 \%$
TS EAMCET 05.08.2021
Thermodynamics
148496
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then
1 compressing the gas through adiabatic process will require more work to be done.
2 compressing the gas isothermally or adiabatically will require the same amount of work.
3 which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.
4 compressing the gas isothermally will require more work to be done.
Explanation:
A Let us consider that the gas undergoes compression from $\mathrm{V}$ to $\frac{\mathrm{V}}{2}$. From $\mathrm{P}-\mathrm{V}$ diagram, when we compare the area under the isothermal process curve and the adiabatic process curve, the work done for adiabatic process is higher than that of work done by isothermal process.
148452
In an adiabatic expansion of an ideal gas the product of pressure and volume :
1 decreases
2 increases
3 remains constant
4 at first increases and then decreases
Explanation:
A Ideal gas equation is given by, $\mathrm{PV}=\mathrm{nRT}$ $\because \mathrm{T}$ is decreasing, $\mathrm{PV}$ also decreases. If $\mathrm{T}$ decreases $\mathrm{PV}$ also decreases because in an adiabatic expansion. Internal energy decreases and temperature also decreases.
Karnataka CET-2020
Thermodynamics
148461
In an adiabatic process
1 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
2 $\mathrm{TV}^{\gamma-1}=$ constant
3 $\mathrm{PV}=$ constant
4 All of these
Explanation:
B For an adiabatic process, $\mathrm{PV}^{\gamma}=$ constant. Using the ideal gas equation and substituting $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$ In the above equation gives $\mathrm{TV}^{\gamma-1}=$ constant Where $\gamma$ is the specific heat ratio
UP CPMT-2003
Thermodynamics
148463
During adiabatic expansion, the increase in the volume is associated with
1 increase in pressure and decrease in $\mathrm{T}$
2 decrease in pressure and increase in $\mathrm{T}$
3 increase in pressure and $\mathrm{T}$
4 decrease in pressure and $\mathrm{T}$
Explanation:
D $\mathrm{PV}^{\gamma}=$ constant And, $\mathrm{TV}^{\gamma-1}=$ constant From the above equation increasing in volume pressure and temperature both are decreases.
TS EAMCET 30.07.2022
Thermodynamics
148464
An ideal gas undergoes an adiabatic process. If the pressure of the gas is reduced by $0.1 \%$ then the volume is changed by $\left(\right.$ Given $\left.\gamma=\frac{C_{p}}{C_{v}}=5 / 3\right)$
1 $0.1 \%$
2 $0.05 \%$
3 $0.06 \%$
4 $-0.05 \%$
Explanation:
C From the relation, $\mathrm{PV}^{\gamma}=\mathrm{K}$ Now differentiating we get, $\mathrm{PV}^{\gamma-1} \mathrm{dV}+\mathrm{dP} \cdot \mathrm{V}^{\gamma}=0$ By making proper rearrangements we get, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{\mathrm{dP}}{\gamma \mathrm{P}}$ In adiabatic process on a gas with $\gamma=1.4$, the pressure of the is reduced by $0.1 \%$. The volume (decreases) changes by about, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{0.1}{1.4} \times 100=0.06 \%$
TS EAMCET 05.08.2021
Thermodynamics
148496
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then
1 compressing the gas through adiabatic process will require more work to be done.
2 compressing the gas isothermally or adiabatically will require the same amount of work.
3 which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.
4 compressing the gas isothermally will require more work to be done.
Explanation:
A Let us consider that the gas undergoes compression from $\mathrm{V}$ to $\frac{\mathrm{V}}{2}$. From $\mathrm{P}-\mathrm{V}$ diagram, when we compare the area under the isothermal process curve and the adiabatic process curve, the work done for adiabatic process is higher than that of work done by isothermal process.
148452
In an adiabatic expansion of an ideal gas the product of pressure and volume :
1 decreases
2 increases
3 remains constant
4 at first increases and then decreases
Explanation:
A Ideal gas equation is given by, $\mathrm{PV}=\mathrm{nRT}$ $\because \mathrm{T}$ is decreasing, $\mathrm{PV}$ also decreases. If $\mathrm{T}$ decreases $\mathrm{PV}$ also decreases because in an adiabatic expansion. Internal energy decreases and temperature also decreases.
Karnataka CET-2020
Thermodynamics
148461
In an adiabatic process
1 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
2 $\mathrm{TV}^{\gamma-1}=$ constant
3 $\mathrm{PV}=$ constant
4 All of these
Explanation:
B For an adiabatic process, $\mathrm{PV}^{\gamma}=$ constant. Using the ideal gas equation and substituting $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$ In the above equation gives $\mathrm{TV}^{\gamma-1}=$ constant Where $\gamma$ is the specific heat ratio
UP CPMT-2003
Thermodynamics
148463
During adiabatic expansion, the increase in the volume is associated with
1 increase in pressure and decrease in $\mathrm{T}$
2 decrease in pressure and increase in $\mathrm{T}$
3 increase in pressure and $\mathrm{T}$
4 decrease in pressure and $\mathrm{T}$
Explanation:
D $\mathrm{PV}^{\gamma}=$ constant And, $\mathrm{TV}^{\gamma-1}=$ constant From the above equation increasing in volume pressure and temperature both are decreases.
TS EAMCET 30.07.2022
Thermodynamics
148464
An ideal gas undergoes an adiabatic process. If the pressure of the gas is reduced by $0.1 \%$ then the volume is changed by $\left(\right.$ Given $\left.\gamma=\frac{C_{p}}{C_{v}}=5 / 3\right)$
1 $0.1 \%$
2 $0.05 \%$
3 $0.06 \%$
4 $-0.05 \%$
Explanation:
C From the relation, $\mathrm{PV}^{\gamma}=\mathrm{K}$ Now differentiating we get, $\mathrm{PV}^{\gamma-1} \mathrm{dV}+\mathrm{dP} \cdot \mathrm{V}^{\gamma}=0$ By making proper rearrangements we get, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{\mathrm{dP}}{\gamma \mathrm{P}}$ In adiabatic process on a gas with $\gamma=1.4$, the pressure of the is reduced by $0.1 \%$. The volume (decreases) changes by about, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{0.1}{1.4} \times 100=0.06 \%$
TS EAMCET 05.08.2021
Thermodynamics
148496
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then
1 compressing the gas through adiabatic process will require more work to be done.
2 compressing the gas isothermally or adiabatically will require the same amount of work.
3 which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.
4 compressing the gas isothermally will require more work to be done.
Explanation:
A Let us consider that the gas undergoes compression from $\mathrm{V}$ to $\frac{\mathrm{V}}{2}$. From $\mathrm{P}-\mathrm{V}$ diagram, when we compare the area under the isothermal process curve and the adiabatic process curve, the work done for adiabatic process is higher than that of work done by isothermal process.
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Thermodynamics
148452
In an adiabatic expansion of an ideal gas the product of pressure and volume :
1 decreases
2 increases
3 remains constant
4 at first increases and then decreases
Explanation:
A Ideal gas equation is given by, $\mathrm{PV}=\mathrm{nRT}$ $\because \mathrm{T}$ is decreasing, $\mathrm{PV}$ also decreases. If $\mathrm{T}$ decreases $\mathrm{PV}$ also decreases because in an adiabatic expansion. Internal energy decreases and temperature also decreases.
Karnataka CET-2020
Thermodynamics
148461
In an adiabatic process
1 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
2 $\mathrm{TV}^{\gamma-1}=$ constant
3 $\mathrm{PV}=$ constant
4 All of these
Explanation:
B For an adiabatic process, $\mathrm{PV}^{\gamma}=$ constant. Using the ideal gas equation and substituting $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$ In the above equation gives $\mathrm{TV}^{\gamma-1}=$ constant Where $\gamma$ is the specific heat ratio
UP CPMT-2003
Thermodynamics
148463
During adiabatic expansion, the increase in the volume is associated with
1 increase in pressure and decrease in $\mathrm{T}$
2 decrease in pressure and increase in $\mathrm{T}$
3 increase in pressure and $\mathrm{T}$
4 decrease in pressure and $\mathrm{T}$
Explanation:
D $\mathrm{PV}^{\gamma}=$ constant And, $\mathrm{TV}^{\gamma-1}=$ constant From the above equation increasing in volume pressure and temperature both are decreases.
TS EAMCET 30.07.2022
Thermodynamics
148464
An ideal gas undergoes an adiabatic process. If the pressure of the gas is reduced by $0.1 \%$ then the volume is changed by $\left(\right.$ Given $\left.\gamma=\frac{C_{p}}{C_{v}}=5 / 3\right)$
1 $0.1 \%$
2 $0.05 \%$
3 $0.06 \%$
4 $-0.05 \%$
Explanation:
C From the relation, $\mathrm{PV}^{\gamma}=\mathrm{K}$ Now differentiating we get, $\mathrm{PV}^{\gamma-1} \mathrm{dV}+\mathrm{dP} \cdot \mathrm{V}^{\gamma}=0$ By making proper rearrangements we get, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{\mathrm{dP}}{\gamma \mathrm{P}}$ In adiabatic process on a gas with $\gamma=1.4$, the pressure of the is reduced by $0.1 \%$. The volume (decreases) changes by about, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{0.1}{1.4} \times 100=0.06 \%$
TS EAMCET 05.08.2021
Thermodynamics
148496
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then
1 compressing the gas through adiabatic process will require more work to be done.
2 compressing the gas isothermally or adiabatically will require the same amount of work.
3 which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.
4 compressing the gas isothermally will require more work to be done.
Explanation:
A Let us consider that the gas undergoes compression from $\mathrm{V}$ to $\frac{\mathrm{V}}{2}$. From $\mathrm{P}-\mathrm{V}$ diagram, when we compare the area under the isothermal process curve and the adiabatic process curve, the work done for adiabatic process is higher than that of work done by isothermal process.
148452
In an adiabatic expansion of an ideal gas the product of pressure and volume :
1 decreases
2 increases
3 remains constant
4 at first increases and then decreases
Explanation:
A Ideal gas equation is given by, $\mathrm{PV}=\mathrm{nRT}$ $\because \mathrm{T}$ is decreasing, $\mathrm{PV}$ also decreases. If $\mathrm{T}$ decreases $\mathrm{PV}$ also decreases because in an adiabatic expansion. Internal energy decreases and temperature also decreases.
Karnataka CET-2020
Thermodynamics
148461
In an adiabatic process
1 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
2 $\mathrm{TV}^{\gamma-1}=$ constant
3 $\mathrm{PV}=$ constant
4 All of these
Explanation:
B For an adiabatic process, $\mathrm{PV}^{\gamma}=$ constant. Using the ideal gas equation and substituting $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$ In the above equation gives $\mathrm{TV}^{\gamma-1}=$ constant Where $\gamma$ is the specific heat ratio
UP CPMT-2003
Thermodynamics
148463
During adiabatic expansion, the increase in the volume is associated with
1 increase in pressure and decrease in $\mathrm{T}$
2 decrease in pressure and increase in $\mathrm{T}$
3 increase in pressure and $\mathrm{T}$
4 decrease in pressure and $\mathrm{T}$
Explanation:
D $\mathrm{PV}^{\gamma}=$ constant And, $\mathrm{TV}^{\gamma-1}=$ constant From the above equation increasing in volume pressure and temperature both are decreases.
TS EAMCET 30.07.2022
Thermodynamics
148464
An ideal gas undergoes an adiabatic process. If the pressure of the gas is reduced by $0.1 \%$ then the volume is changed by $\left(\right.$ Given $\left.\gamma=\frac{C_{p}}{C_{v}}=5 / 3\right)$
1 $0.1 \%$
2 $0.05 \%$
3 $0.06 \%$
4 $-0.05 \%$
Explanation:
C From the relation, $\mathrm{PV}^{\gamma}=\mathrm{K}$ Now differentiating we get, $\mathrm{PV}^{\gamma-1} \mathrm{dV}+\mathrm{dP} \cdot \mathrm{V}^{\gamma}=0$ By making proper rearrangements we get, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{\mathrm{dP}}{\gamma \mathrm{P}}$ In adiabatic process on a gas with $\gamma=1.4$, the pressure of the is reduced by $0.1 \%$. The volume (decreases) changes by about, $\frac{-\mathrm{dV}}{\mathrm{V}}=\frac{0.1}{1.4} \times 100=0.06 \%$
TS EAMCET 05.08.2021
Thermodynamics
148496
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then
1 compressing the gas through adiabatic process will require more work to be done.
2 compressing the gas isothermally or adiabatically will require the same amount of work.
3 which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.
4 compressing the gas isothermally will require more work to be done.
Explanation:
A Let us consider that the gas undergoes compression from $\mathrm{V}$ to $\frac{\mathrm{V}}{2}$. From $\mathrm{P}-\mathrm{V}$ diagram, when we compare the area under the isothermal process curve and the adiabatic process curve, the work done for adiabatic process is higher than that of work done by isothermal process.
