NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
148488
during adiabatic expansion of 2 moles of a gas, the change in internal energy was found to be equal to $100 \mathrm{~J}$. The work done during the process will be equal to
1 Zero
2 $200 \mathrm{~J}$
3 $-100 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
D according to first law $\Delta \mathrm{Q}=\mathrm{W}+\Delta \mathrm{U}$ during adiabatic expansion $\{\Delta \mathrm{Q}=0\}$ $\mathrm{W}+\Delta \mathrm{U}=0$ $\mathrm{W}=-\Delta \mathrm{U}$ $\mathrm{W}=-(-100)$ $\mathrm{W}=100 \mathrm{~J}$
AP EAMCET-25.09.2020
Thermodynamics
148489
When the ideal gas is compressed isothermally, then its pressure
1 Increases
2 Decreases
3 remains the same
4 first increases and then decreases
Explanation:
A When the ideal gas is compressed isothermally $(\mathrm{T}=\mathrm{C})$ Then the volume of gas gets decreases As we know, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{T}=\mathrm{C}$ $\mathrm{PV}=\text { Constant }$ $\mathrm{P} \propto \frac{1}{\mathrm{~V}}$ Volume decreases, when compressed isothermally then pressure will increase. Hence, it is said that when ideal gas is compressed isothermally the pressure will increase.
AP EAMCET-24.08.2021
Thermodynamics
148490
A given system undergoes a change in which the work done by the system equals the decrease in its internal energy. The system must have undergone an
1 Isothermal change
2 Adiabatic change
3 Isobaric change
4 Isochoric change
Explanation:
B We know that, $\Delta \mathrm{W}+\Delta \mathrm{U}=0$ $\Delta \mathrm{W}=-\Delta \mathrm{U}$ According to first law of thermodynamics $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=0$ ( $\because$ adiabatic process) $\Delta \mathrm{U}=-\Delta \mathrm{W}$ $\Delta \mathrm{W}=-\Delta \mathrm{U}$ Hence, process is adiabatic.
AP EAMCET-25.08.2021
Thermodynamics
148491
The mole of an ideal gas expands adiabatically from $200 \mathrm{~K}$ to $250 \mathrm{~K}$. If the specific heat of the gas at constant volume is $0.8 \mathrm{~kJ} \mathrm{~kg}^{-1}$, then the work done by the gas is
1 $20 \mathrm{~J}$
2 $20 \mathrm{~kJ}$
3 $40 \mathrm{~J}$
4 $40 \mathrm{~kJ}$
Explanation:
D Given, Temperature change $=250 \mathrm{~K}-200 \mathrm{~K}=50 \mathrm{~K}$ Specific heat $=0.8 \mathrm{~kJ} / \mathrm{kg}$ $\mathrm{dU}=\mathrm{ms} \Delta \mathrm{T}$ $\mathrm{dU}=1 \times 0.8 \times 50$ $\mathrm{dU}=40 \mathrm{~kJ}$ As we know that, $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dW}=-\mathrm{dU}$ $\mathrm{dW}=-40 \quad[\mathrm{dQ}=0 \text { for adiabatic process }]$ Work done by the gas $=40 \mathrm{~kJ}$.
148488
during adiabatic expansion of 2 moles of a gas, the change in internal energy was found to be equal to $100 \mathrm{~J}$. The work done during the process will be equal to
1 Zero
2 $200 \mathrm{~J}$
3 $-100 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
D according to first law $\Delta \mathrm{Q}=\mathrm{W}+\Delta \mathrm{U}$ during adiabatic expansion $\{\Delta \mathrm{Q}=0\}$ $\mathrm{W}+\Delta \mathrm{U}=0$ $\mathrm{W}=-\Delta \mathrm{U}$ $\mathrm{W}=-(-100)$ $\mathrm{W}=100 \mathrm{~J}$
AP EAMCET-25.09.2020
Thermodynamics
148489
When the ideal gas is compressed isothermally, then its pressure
1 Increases
2 Decreases
3 remains the same
4 first increases and then decreases
Explanation:
A When the ideal gas is compressed isothermally $(\mathrm{T}=\mathrm{C})$ Then the volume of gas gets decreases As we know, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{T}=\mathrm{C}$ $\mathrm{PV}=\text { Constant }$ $\mathrm{P} \propto \frac{1}{\mathrm{~V}}$ Volume decreases, when compressed isothermally then pressure will increase. Hence, it is said that when ideal gas is compressed isothermally the pressure will increase.
AP EAMCET-24.08.2021
Thermodynamics
148490
A given system undergoes a change in which the work done by the system equals the decrease in its internal energy. The system must have undergone an
1 Isothermal change
2 Adiabatic change
3 Isobaric change
4 Isochoric change
Explanation:
B We know that, $\Delta \mathrm{W}+\Delta \mathrm{U}=0$ $\Delta \mathrm{W}=-\Delta \mathrm{U}$ According to first law of thermodynamics $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=0$ ( $\because$ adiabatic process) $\Delta \mathrm{U}=-\Delta \mathrm{W}$ $\Delta \mathrm{W}=-\Delta \mathrm{U}$ Hence, process is adiabatic.
AP EAMCET-25.08.2021
Thermodynamics
148491
The mole of an ideal gas expands adiabatically from $200 \mathrm{~K}$ to $250 \mathrm{~K}$. If the specific heat of the gas at constant volume is $0.8 \mathrm{~kJ} \mathrm{~kg}^{-1}$, then the work done by the gas is
1 $20 \mathrm{~J}$
2 $20 \mathrm{~kJ}$
3 $40 \mathrm{~J}$
4 $40 \mathrm{~kJ}$
Explanation:
D Given, Temperature change $=250 \mathrm{~K}-200 \mathrm{~K}=50 \mathrm{~K}$ Specific heat $=0.8 \mathrm{~kJ} / \mathrm{kg}$ $\mathrm{dU}=\mathrm{ms} \Delta \mathrm{T}$ $\mathrm{dU}=1 \times 0.8 \times 50$ $\mathrm{dU}=40 \mathrm{~kJ}$ As we know that, $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dW}=-\mathrm{dU}$ $\mathrm{dW}=-40 \quad[\mathrm{dQ}=0 \text { for adiabatic process }]$ Work done by the gas $=40 \mathrm{~kJ}$.
148488
during adiabatic expansion of 2 moles of a gas, the change in internal energy was found to be equal to $100 \mathrm{~J}$. The work done during the process will be equal to
1 Zero
2 $200 \mathrm{~J}$
3 $-100 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
D according to first law $\Delta \mathrm{Q}=\mathrm{W}+\Delta \mathrm{U}$ during adiabatic expansion $\{\Delta \mathrm{Q}=0\}$ $\mathrm{W}+\Delta \mathrm{U}=0$ $\mathrm{W}=-\Delta \mathrm{U}$ $\mathrm{W}=-(-100)$ $\mathrm{W}=100 \mathrm{~J}$
AP EAMCET-25.09.2020
Thermodynamics
148489
When the ideal gas is compressed isothermally, then its pressure
1 Increases
2 Decreases
3 remains the same
4 first increases and then decreases
Explanation:
A When the ideal gas is compressed isothermally $(\mathrm{T}=\mathrm{C})$ Then the volume of gas gets decreases As we know, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{T}=\mathrm{C}$ $\mathrm{PV}=\text { Constant }$ $\mathrm{P} \propto \frac{1}{\mathrm{~V}}$ Volume decreases, when compressed isothermally then pressure will increase. Hence, it is said that when ideal gas is compressed isothermally the pressure will increase.
AP EAMCET-24.08.2021
Thermodynamics
148490
A given system undergoes a change in which the work done by the system equals the decrease in its internal energy. The system must have undergone an
1 Isothermal change
2 Adiabatic change
3 Isobaric change
4 Isochoric change
Explanation:
B We know that, $\Delta \mathrm{W}+\Delta \mathrm{U}=0$ $\Delta \mathrm{W}=-\Delta \mathrm{U}$ According to first law of thermodynamics $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=0$ ( $\because$ adiabatic process) $\Delta \mathrm{U}=-\Delta \mathrm{W}$ $\Delta \mathrm{W}=-\Delta \mathrm{U}$ Hence, process is adiabatic.
AP EAMCET-25.08.2021
Thermodynamics
148491
The mole of an ideal gas expands adiabatically from $200 \mathrm{~K}$ to $250 \mathrm{~K}$. If the specific heat of the gas at constant volume is $0.8 \mathrm{~kJ} \mathrm{~kg}^{-1}$, then the work done by the gas is
1 $20 \mathrm{~J}$
2 $20 \mathrm{~kJ}$
3 $40 \mathrm{~J}$
4 $40 \mathrm{~kJ}$
Explanation:
D Given, Temperature change $=250 \mathrm{~K}-200 \mathrm{~K}=50 \mathrm{~K}$ Specific heat $=0.8 \mathrm{~kJ} / \mathrm{kg}$ $\mathrm{dU}=\mathrm{ms} \Delta \mathrm{T}$ $\mathrm{dU}=1 \times 0.8 \times 50$ $\mathrm{dU}=40 \mathrm{~kJ}$ As we know that, $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dW}=-\mathrm{dU}$ $\mathrm{dW}=-40 \quad[\mathrm{dQ}=0 \text { for adiabatic process }]$ Work done by the gas $=40 \mathrm{~kJ}$.
148488
during adiabatic expansion of 2 moles of a gas, the change in internal energy was found to be equal to $100 \mathrm{~J}$. The work done during the process will be equal to
1 Zero
2 $200 \mathrm{~J}$
3 $-100 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
D according to first law $\Delta \mathrm{Q}=\mathrm{W}+\Delta \mathrm{U}$ during adiabatic expansion $\{\Delta \mathrm{Q}=0\}$ $\mathrm{W}+\Delta \mathrm{U}=0$ $\mathrm{W}=-\Delta \mathrm{U}$ $\mathrm{W}=-(-100)$ $\mathrm{W}=100 \mathrm{~J}$
AP EAMCET-25.09.2020
Thermodynamics
148489
When the ideal gas is compressed isothermally, then its pressure
1 Increases
2 Decreases
3 remains the same
4 first increases and then decreases
Explanation:
A When the ideal gas is compressed isothermally $(\mathrm{T}=\mathrm{C})$ Then the volume of gas gets decreases As we know, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{T}=\mathrm{C}$ $\mathrm{PV}=\text { Constant }$ $\mathrm{P} \propto \frac{1}{\mathrm{~V}}$ Volume decreases, when compressed isothermally then pressure will increase. Hence, it is said that when ideal gas is compressed isothermally the pressure will increase.
AP EAMCET-24.08.2021
Thermodynamics
148490
A given system undergoes a change in which the work done by the system equals the decrease in its internal energy. The system must have undergone an
1 Isothermal change
2 Adiabatic change
3 Isobaric change
4 Isochoric change
Explanation:
B We know that, $\Delta \mathrm{W}+\Delta \mathrm{U}=0$ $\Delta \mathrm{W}=-\Delta \mathrm{U}$ According to first law of thermodynamics $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=0$ ( $\because$ adiabatic process) $\Delta \mathrm{U}=-\Delta \mathrm{W}$ $\Delta \mathrm{W}=-\Delta \mathrm{U}$ Hence, process is adiabatic.
AP EAMCET-25.08.2021
Thermodynamics
148491
The mole of an ideal gas expands adiabatically from $200 \mathrm{~K}$ to $250 \mathrm{~K}$. If the specific heat of the gas at constant volume is $0.8 \mathrm{~kJ} \mathrm{~kg}^{-1}$, then the work done by the gas is
1 $20 \mathrm{~J}$
2 $20 \mathrm{~kJ}$
3 $40 \mathrm{~J}$
4 $40 \mathrm{~kJ}$
Explanation:
D Given, Temperature change $=250 \mathrm{~K}-200 \mathrm{~K}=50 \mathrm{~K}$ Specific heat $=0.8 \mathrm{~kJ} / \mathrm{kg}$ $\mathrm{dU}=\mathrm{ms} \Delta \mathrm{T}$ $\mathrm{dU}=1 \times 0.8 \times 50$ $\mathrm{dU}=40 \mathrm{~kJ}$ As we know that, $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dW}=-\mathrm{dU}$ $\mathrm{dW}=-40 \quad[\mathrm{dQ}=0 \text { for adiabatic process }]$ Work done by the gas $=40 \mathrm{~kJ}$.
