148410
An ideal gas undergoes an isothermal change in volume with pressure, then
1 $\mathrm{PV}=$ constant
2 $(\mathrm{PV})^{\gamma}=$ constant
3 $\mathrm{PV}^{\gamma}=$ constant
4 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
Explanation:
A For an ideal gas, the product of pressure and volume ( $\mathrm{P}-\mathrm{V})$ is a constant, if the gas is kept in isothermal condition. In other words, From ideal gas law, PV $=\mathrm{nRT}$. $\mathrm{PV}=\mathrm{nRT}=\text { constant }$ $\therefore \quad \mathrm{PV}=$ constant
J and K CET- 2001
Thermodynamics
148412
Name the process in which Boyle's law is applicable?
1 Adiabatic process
2 Isochoric process
3 Isobaric process
4 Isothermal process
Explanation:
D Boyle's law is applicable to an isothermal process where temperature remains constant. Boyle's law, Temperature $=$ constant
J and K-CET-2019
Thermodynamics
148417
In isothermal process, which of the following is not true?
1 Temperature remains constant
2 Internal energy does not change
3 No heat enters or leaves the system
4 none of the above
Explanation:
C We know that, In an isothermal process, Temperature is constant i.e. $\mathrm{PV}=$ constant According to joule's law, internal energy remains constant for isothermal process. In isothermal process, heat can enters or leaves the system, to keep the temperature constant. So, statement (c) is wrong.
JIPMER-2005
Thermodynamics
148435
One mole of an ideal gas undergoes an isothermal change at temperature $T$, so that its, volume $V$ is doubled, $R$ is the molar gas constant. Work done by the gas during this changes is
1 RT $\log 4$
2 $\mathrm{RT} \log 2$
3 RT $\log 1$
4 $\mathrm{RT} \log 3$
Explanation:
B Given, $\mathrm{V}_{\mathrm{i}}=\mathrm{V}, \mathrm{V}_{\mathrm{f}}=2 \mathrm{~V}$ So, under isothermal process work is given by the relation $\mathrm{W}=\mathrm{RT} \log \left(\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right)$ $\mathrm{W}=\mathrm{RT} \log \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{W}=\mathrm{RT} \log (2)$
AP EMCET(Medical)-2008
Thermodynamics
148429
Which of the following, in general, is a slow process?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
A Slow process is generally prefer to isothermal process. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. i.e. $\Delta \mathrm{T}=0$
148410
An ideal gas undergoes an isothermal change in volume with pressure, then
1 $\mathrm{PV}=$ constant
2 $(\mathrm{PV})^{\gamma}=$ constant
3 $\mathrm{PV}^{\gamma}=$ constant
4 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
Explanation:
A For an ideal gas, the product of pressure and volume ( $\mathrm{P}-\mathrm{V})$ is a constant, if the gas is kept in isothermal condition. In other words, From ideal gas law, PV $=\mathrm{nRT}$. $\mathrm{PV}=\mathrm{nRT}=\text { constant }$ $\therefore \quad \mathrm{PV}=$ constant
J and K CET- 2001
Thermodynamics
148412
Name the process in which Boyle's law is applicable?
1 Adiabatic process
2 Isochoric process
3 Isobaric process
4 Isothermal process
Explanation:
D Boyle's law is applicable to an isothermal process where temperature remains constant. Boyle's law, Temperature $=$ constant
J and K-CET-2019
Thermodynamics
148417
In isothermal process, which of the following is not true?
1 Temperature remains constant
2 Internal energy does not change
3 No heat enters or leaves the system
4 none of the above
Explanation:
C We know that, In an isothermal process, Temperature is constant i.e. $\mathrm{PV}=$ constant According to joule's law, internal energy remains constant for isothermal process. In isothermal process, heat can enters or leaves the system, to keep the temperature constant. So, statement (c) is wrong.
JIPMER-2005
Thermodynamics
148435
One mole of an ideal gas undergoes an isothermal change at temperature $T$, so that its, volume $V$ is doubled, $R$ is the molar gas constant. Work done by the gas during this changes is
1 RT $\log 4$
2 $\mathrm{RT} \log 2$
3 RT $\log 1$
4 $\mathrm{RT} \log 3$
Explanation:
B Given, $\mathrm{V}_{\mathrm{i}}=\mathrm{V}, \mathrm{V}_{\mathrm{f}}=2 \mathrm{~V}$ So, under isothermal process work is given by the relation $\mathrm{W}=\mathrm{RT} \log \left(\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right)$ $\mathrm{W}=\mathrm{RT} \log \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{W}=\mathrm{RT} \log (2)$
AP EMCET(Medical)-2008
Thermodynamics
148429
Which of the following, in general, is a slow process?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
A Slow process is generally prefer to isothermal process. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. i.e. $\Delta \mathrm{T}=0$
148410
An ideal gas undergoes an isothermal change in volume with pressure, then
1 $\mathrm{PV}=$ constant
2 $(\mathrm{PV})^{\gamma}=$ constant
3 $\mathrm{PV}^{\gamma}=$ constant
4 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
Explanation:
A For an ideal gas, the product of pressure and volume ( $\mathrm{P}-\mathrm{V})$ is a constant, if the gas is kept in isothermal condition. In other words, From ideal gas law, PV $=\mathrm{nRT}$. $\mathrm{PV}=\mathrm{nRT}=\text { constant }$ $\therefore \quad \mathrm{PV}=$ constant
J and K CET- 2001
Thermodynamics
148412
Name the process in which Boyle's law is applicable?
1 Adiabatic process
2 Isochoric process
3 Isobaric process
4 Isothermal process
Explanation:
D Boyle's law is applicable to an isothermal process where temperature remains constant. Boyle's law, Temperature $=$ constant
J and K-CET-2019
Thermodynamics
148417
In isothermal process, which of the following is not true?
1 Temperature remains constant
2 Internal energy does not change
3 No heat enters or leaves the system
4 none of the above
Explanation:
C We know that, In an isothermal process, Temperature is constant i.e. $\mathrm{PV}=$ constant According to joule's law, internal energy remains constant for isothermal process. In isothermal process, heat can enters or leaves the system, to keep the temperature constant. So, statement (c) is wrong.
JIPMER-2005
Thermodynamics
148435
One mole of an ideal gas undergoes an isothermal change at temperature $T$, so that its, volume $V$ is doubled, $R$ is the molar gas constant. Work done by the gas during this changes is
1 RT $\log 4$
2 $\mathrm{RT} \log 2$
3 RT $\log 1$
4 $\mathrm{RT} \log 3$
Explanation:
B Given, $\mathrm{V}_{\mathrm{i}}=\mathrm{V}, \mathrm{V}_{\mathrm{f}}=2 \mathrm{~V}$ So, under isothermal process work is given by the relation $\mathrm{W}=\mathrm{RT} \log \left(\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right)$ $\mathrm{W}=\mathrm{RT} \log \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{W}=\mathrm{RT} \log (2)$
AP EMCET(Medical)-2008
Thermodynamics
148429
Which of the following, in general, is a slow process?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
A Slow process is generally prefer to isothermal process. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. i.e. $\Delta \mathrm{T}=0$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
148410
An ideal gas undergoes an isothermal change in volume with pressure, then
1 $\mathrm{PV}=$ constant
2 $(\mathrm{PV})^{\gamma}=$ constant
3 $\mathrm{PV}^{\gamma}=$ constant
4 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
Explanation:
A For an ideal gas, the product of pressure and volume ( $\mathrm{P}-\mathrm{V})$ is a constant, if the gas is kept in isothermal condition. In other words, From ideal gas law, PV $=\mathrm{nRT}$. $\mathrm{PV}=\mathrm{nRT}=\text { constant }$ $\therefore \quad \mathrm{PV}=$ constant
J and K CET- 2001
Thermodynamics
148412
Name the process in which Boyle's law is applicable?
1 Adiabatic process
2 Isochoric process
3 Isobaric process
4 Isothermal process
Explanation:
D Boyle's law is applicable to an isothermal process where temperature remains constant. Boyle's law, Temperature $=$ constant
J and K-CET-2019
Thermodynamics
148417
In isothermal process, which of the following is not true?
1 Temperature remains constant
2 Internal energy does not change
3 No heat enters or leaves the system
4 none of the above
Explanation:
C We know that, In an isothermal process, Temperature is constant i.e. $\mathrm{PV}=$ constant According to joule's law, internal energy remains constant for isothermal process. In isothermal process, heat can enters or leaves the system, to keep the temperature constant. So, statement (c) is wrong.
JIPMER-2005
Thermodynamics
148435
One mole of an ideal gas undergoes an isothermal change at temperature $T$, so that its, volume $V$ is doubled, $R$ is the molar gas constant. Work done by the gas during this changes is
1 RT $\log 4$
2 $\mathrm{RT} \log 2$
3 RT $\log 1$
4 $\mathrm{RT} \log 3$
Explanation:
B Given, $\mathrm{V}_{\mathrm{i}}=\mathrm{V}, \mathrm{V}_{\mathrm{f}}=2 \mathrm{~V}$ So, under isothermal process work is given by the relation $\mathrm{W}=\mathrm{RT} \log \left(\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right)$ $\mathrm{W}=\mathrm{RT} \log \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{W}=\mathrm{RT} \log (2)$
AP EMCET(Medical)-2008
Thermodynamics
148429
Which of the following, in general, is a slow process?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
A Slow process is generally prefer to isothermal process. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. i.e. $\Delta \mathrm{T}=0$
148410
An ideal gas undergoes an isothermal change in volume with pressure, then
1 $\mathrm{PV}=$ constant
2 $(\mathrm{PV})^{\gamma}=$ constant
3 $\mathrm{PV}^{\gamma}=$ constant
4 $\mathrm{P}^{\gamma} \mathrm{V}=$ constant
Explanation:
A For an ideal gas, the product of pressure and volume ( $\mathrm{P}-\mathrm{V})$ is a constant, if the gas is kept in isothermal condition. In other words, From ideal gas law, PV $=\mathrm{nRT}$. $\mathrm{PV}=\mathrm{nRT}=\text { constant }$ $\therefore \quad \mathrm{PV}=$ constant
J and K CET- 2001
Thermodynamics
148412
Name the process in which Boyle's law is applicable?
1 Adiabatic process
2 Isochoric process
3 Isobaric process
4 Isothermal process
Explanation:
D Boyle's law is applicable to an isothermal process where temperature remains constant. Boyle's law, Temperature $=$ constant
J and K-CET-2019
Thermodynamics
148417
In isothermal process, which of the following is not true?
1 Temperature remains constant
2 Internal energy does not change
3 No heat enters or leaves the system
4 none of the above
Explanation:
C We know that, In an isothermal process, Temperature is constant i.e. $\mathrm{PV}=$ constant According to joule's law, internal energy remains constant for isothermal process. In isothermal process, heat can enters or leaves the system, to keep the temperature constant. So, statement (c) is wrong.
JIPMER-2005
Thermodynamics
148435
One mole of an ideal gas undergoes an isothermal change at temperature $T$, so that its, volume $V$ is doubled, $R$ is the molar gas constant. Work done by the gas during this changes is
1 RT $\log 4$
2 $\mathrm{RT} \log 2$
3 RT $\log 1$
4 $\mathrm{RT} \log 3$
Explanation:
B Given, $\mathrm{V}_{\mathrm{i}}=\mathrm{V}, \mathrm{V}_{\mathrm{f}}=2 \mathrm{~V}$ So, under isothermal process work is given by the relation $\mathrm{W}=\mathrm{RT} \log \left(\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right)$ $\mathrm{W}=\mathrm{RT} \log \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)$ $\mathrm{W}=\mathrm{RT} \log (2)$
AP EMCET(Medical)-2008
Thermodynamics
148429
Which of the following, in general, is a slow process?
1 Isothermal
2 Adiabatic
3 Isobaric
4 Isochoric
Explanation:
A Slow process is generally prefer to isothermal process. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. i.e. $\Delta \mathrm{T}=0$
