272662
Based on the first law of thermodynamics, which one of the following is correct?
1 For an isochoric process, $\Delta \mathrm{u}=-\mathrm{q}$
2 For an adiabatic process, $\mathrm{q}=0, \Delta \mathrm{U}=\mathrm{W}$
3 For an isothermal process, $\Delta \mathrm{U}=0, \mathrm{q}=-\mathrm{W}$
4 For a cyclic process, $\Delta \mathrm{U}=0$ $\therefore \quad \Delta \mathrm{U}=0=\mathrm{q}+\mathrm{W}$ or $\mathrm{q}=-\mathrm{W}$
Explanation:
(a) For an isochoric process, Work done by the gas $=-\mathrm{P} \Delta \mathrm{V}$ $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\Delta \mathrm{U}=\mathrm{q}-\mathrm{P} \Delta \mathrm{V}$ When, $\Delta \mathrm{V}=0, \Delta \mathrm{U}=\mathrm{q}$ or $\mathrm{qv}$ (b.) For an adiabatic process, $\mathrm{q}=0, \Delta \mathrm{U}=\mathrm{W}$ (c.) For an isothermal process, $\Delta \mathrm{U}=0, \mathrm{q}=-\mathrm{W}$ (d.) For a cyclic process, $\Delta \mathrm{U}=0$ $\therefore \quad \Delta \mathrm{U}=0=\mathrm{q}+\mathrm{W}$ or $\mathrm{q}=-\mathrm{W}$
**[COMEDK-2019]
Thermodynamics
272663
Assertion: In a process, if work $=0$ then $\Delta \mathrm{U}=$ q. Reason: $q$ is difference between initial state and final state of a system.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
At in a process At $\mathrm{W}=0$ From thermodynamics $I^{\text {st }}$ Law $\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}$ At $\mathrm{W}=0$ $\Delta \mathrm{U}=\mathrm{q}$ $\because \quad$ So, Assertion is correct $\mathrm{q} \rightarrow$ path function does not depend on initial and final state at is a path dependent
AIIMS 26 May 2019 (Evening)
Thermodynamics
272664
I mole of an ideal gas expands isothermally and reversibly from 2 lit to 4 lit and 3 moles of same gas expand from 2 lit to $x$ lit and doing same work, what is ' $x$ '?
272665
One mono-atomic gas is expanded adiabatically from $2 \mathrm{~L}$ to $10 \mathrm{~L}$ at $1 \mathrm{~atm}$ external pressure, find $\Delta \mathrm{U}$ (in atm L)?
1 -8
2 0
3 -66.7
4 58.2
Explanation:
Given that, $V_1=2 \mathrm{~L}, \mathrm{~V}_2=10 \mathrm{~L}, \mathrm{P}=1 \mathrm{~atm}$ At adiabatic process $Q=$ constant Process is adiabatic $\therefore \Delta Q=0$ $\therefore \quad \Delta \mathrm{U}=\mathrm{W} =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}$ $=-1(10-2) \mathrm{atm} \mathrm{L}$ $=-8 \mathrm{~atm} \mathrm{~L}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
272662
Based on the first law of thermodynamics, which one of the following is correct?
1 For an isochoric process, $\Delta \mathrm{u}=-\mathrm{q}$
2 For an adiabatic process, $\mathrm{q}=0, \Delta \mathrm{U}=\mathrm{W}$
3 For an isothermal process, $\Delta \mathrm{U}=0, \mathrm{q}=-\mathrm{W}$
4 For a cyclic process, $\Delta \mathrm{U}=0$ $\therefore \quad \Delta \mathrm{U}=0=\mathrm{q}+\mathrm{W}$ or $\mathrm{q}=-\mathrm{W}$
Explanation:
(a) For an isochoric process, Work done by the gas $=-\mathrm{P} \Delta \mathrm{V}$ $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\Delta \mathrm{U}=\mathrm{q}-\mathrm{P} \Delta \mathrm{V}$ When, $\Delta \mathrm{V}=0, \Delta \mathrm{U}=\mathrm{q}$ or $\mathrm{qv}$ (b.) For an adiabatic process, $\mathrm{q}=0, \Delta \mathrm{U}=\mathrm{W}$ (c.) For an isothermal process, $\Delta \mathrm{U}=0, \mathrm{q}=-\mathrm{W}$ (d.) For a cyclic process, $\Delta \mathrm{U}=0$ $\therefore \quad \Delta \mathrm{U}=0=\mathrm{q}+\mathrm{W}$ or $\mathrm{q}=-\mathrm{W}$
**[COMEDK-2019]
Thermodynamics
272663
Assertion: In a process, if work $=0$ then $\Delta \mathrm{U}=$ q. Reason: $q$ is difference between initial state and final state of a system.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
At in a process At $\mathrm{W}=0$ From thermodynamics $I^{\text {st }}$ Law $\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}$ At $\mathrm{W}=0$ $\Delta \mathrm{U}=\mathrm{q}$ $\because \quad$ So, Assertion is correct $\mathrm{q} \rightarrow$ path function does not depend on initial and final state at is a path dependent
AIIMS 26 May 2019 (Evening)
Thermodynamics
272664
I mole of an ideal gas expands isothermally and reversibly from 2 lit to 4 lit and 3 moles of same gas expand from 2 lit to $x$ lit and doing same work, what is ' $x$ '?
272665
One mono-atomic gas is expanded adiabatically from $2 \mathrm{~L}$ to $10 \mathrm{~L}$ at $1 \mathrm{~atm}$ external pressure, find $\Delta \mathrm{U}$ (in atm L)?
1 -8
2 0
3 -66.7
4 58.2
Explanation:
Given that, $V_1=2 \mathrm{~L}, \mathrm{~V}_2=10 \mathrm{~L}, \mathrm{P}=1 \mathrm{~atm}$ At adiabatic process $Q=$ constant Process is adiabatic $\therefore \Delta Q=0$ $\therefore \quad \Delta \mathrm{U}=\mathrm{W} =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}$ $=-1(10-2) \mathrm{atm} \mathrm{L}$ $=-8 \mathrm{~atm} \mathrm{~L}$
272662
Based on the first law of thermodynamics, which one of the following is correct?
1 For an isochoric process, $\Delta \mathrm{u}=-\mathrm{q}$
2 For an adiabatic process, $\mathrm{q}=0, \Delta \mathrm{U}=\mathrm{W}$
3 For an isothermal process, $\Delta \mathrm{U}=0, \mathrm{q}=-\mathrm{W}$
4 For a cyclic process, $\Delta \mathrm{U}=0$ $\therefore \quad \Delta \mathrm{U}=0=\mathrm{q}+\mathrm{W}$ or $\mathrm{q}=-\mathrm{W}$
Explanation:
(a) For an isochoric process, Work done by the gas $=-\mathrm{P} \Delta \mathrm{V}$ $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\Delta \mathrm{U}=\mathrm{q}-\mathrm{P} \Delta \mathrm{V}$ When, $\Delta \mathrm{V}=0, \Delta \mathrm{U}=\mathrm{q}$ or $\mathrm{qv}$ (b.) For an adiabatic process, $\mathrm{q}=0, \Delta \mathrm{U}=\mathrm{W}$ (c.) For an isothermal process, $\Delta \mathrm{U}=0, \mathrm{q}=-\mathrm{W}$ (d.) For a cyclic process, $\Delta \mathrm{U}=0$ $\therefore \quad \Delta \mathrm{U}=0=\mathrm{q}+\mathrm{W}$ or $\mathrm{q}=-\mathrm{W}$
**[COMEDK-2019]
Thermodynamics
272663
Assertion: In a process, if work $=0$ then $\Delta \mathrm{U}=$ q. Reason: $q$ is difference between initial state and final state of a system.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
At in a process At $\mathrm{W}=0$ From thermodynamics $I^{\text {st }}$ Law $\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}$ At $\mathrm{W}=0$ $\Delta \mathrm{U}=\mathrm{q}$ $\because \quad$ So, Assertion is correct $\mathrm{q} \rightarrow$ path function does not depend on initial and final state at is a path dependent
AIIMS 26 May 2019 (Evening)
Thermodynamics
272664
I mole of an ideal gas expands isothermally and reversibly from 2 lit to 4 lit and 3 moles of same gas expand from 2 lit to $x$ lit and doing same work, what is ' $x$ '?
272665
One mono-atomic gas is expanded adiabatically from $2 \mathrm{~L}$ to $10 \mathrm{~L}$ at $1 \mathrm{~atm}$ external pressure, find $\Delta \mathrm{U}$ (in atm L)?
1 -8
2 0
3 -66.7
4 58.2
Explanation:
Given that, $V_1=2 \mathrm{~L}, \mathrm{~V}_2=10 \mathrm{~L}, \mathrm{P}=1 \mathrm{~atm}$ At adiabatic process $Q=$ constant Process is adiabatic $\therefore \Delta Q=0$ $\therefore \quad \Delta \mathrm{U}=\mathrm{W} =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}$ $=-1(10-2) \mathrm{atm} \mathrm{L}$ $=-8 \mathrm{~atm} \mathrm{~L}$
272662
Based on the first law of thermodynamics, which one of the following is correct?
1 For an isochoric process, $\Delta \mathrm{u}=-\mathrm{q}$
2 For an adiabatic process, $\mathrm{q}=0, \Delta \mathrm{U}=\mathrm{W}$
3 For an isothermal process, $\Delta \mathrm{U}=0, \mathrm{q}=-\mathrm{W}$
4 For a cyclic process, $\Delta \mathrm{U}=0$ $\therefore \quad \Delta \mathrm{U}=0=\mathrm{q}+\mathrm{W}$ or $\mathrm{q}=-\mathrm{W}$
Explanation:
(a) For an isochoric process, Work done by the gas $=-\mathrm{P} \Delta \mathrm{V}$ $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\Delta \mathrm{U}=\mathrm{q}-\mathrm{P} \Delta \mathrm{V}$ When, $\Delta \mathrm{V}=0, \Delta \mathrm{U}=\mathrm{q}$ or $\mathrm{qv}$ (b.) For an adiabatic process, $\mathrm{q}=0, \Delta \mathrm{U}=\mathrm{W}$ (c.) For an isothermal process, $\Delta \mathrm{U}=0, \mathrm{q}=-\mathrm{W}$ (d.) For a cyclic process, $\Delta \mathrm{U}=0$ $\therefore \quad \Delta \mathrm{U}=0=\mathrm{q}+\mathrm{W}$ or $\mathrm{q}=-\mathrm{W}$
**[COMEDK-2019]
Thermodynamics
272663
Assertion: In a process, if work $=0$ then $\Delta \mathrm{U}=$ q. Reason: $q$ is difference between initial state and final state of a system.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
At in a process At $\mathrm{W}=0$ From thermodynamics $I^{\text {st }}$ Law $\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}$ At $\mathrm{W}=0$ $\Delta \mathrm{U}=\mathrm{q}$ $\because \quad$ So, Assertion is correct $\mathrm{q} \rightarrow$ path function does not depend on initial and final state at is a path dependent
AIIMS 26 May 2019 (Evening)
Thermodynamics
272664
I mole of an ideal gas expands isothermally and reversibly from 2 lit to 4 lit and 3 moles of same gas expand from 2 lit to $x$ lit and doing same work, what is ' $x$ '?
272665
One mono-atomic gas is expanded adiabatically from $2 \mathrm{~L}$ to $10 \mathrm{~L}$ at $1 \mathrm{~atm}$ external pressure, find $\Delta \mathrm{U}$ (in atm L)?
1 -8
2 0
3 -66.7
4 58.2
Explanation:
Given that, $V_1=2 \mathrm{~L}, \mathrm{~V}_2=10 \mathrm{~L}, \mathrm{P}=1 \mathrm{~atm}$ At adiabatic process $Q=$ constant Process is adiabatic $\therefore \Delta Q=0$ $\therefore \quad \Delta \mathrm{U}=\mathrm{W} =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}$ $=-1(10-2) \mathrm{atm} \mathrm{L}$ $=-8 \mathrm{~atm} \mathrm{~L}$