NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
148639
A Carnot engine whose low-temperature reservoir is at $27^{\circ} \mathrm{C}$, has an efficiency $37.5 \%$. The high temperature reservoir is at
1 $480{ }^{\circ} \mathrm{C}$
2 $327^{\circ} \mathrm{C}$
3 $307^{\circ} \mathrm{C}$
4 $207^{\circ} \mathrm{C}$
Explanation:
D Given, $\mathrm{T}_{2}=27+273=300 \mathrm{~K}$ and efficiency $(\eta)=37.5 \%=\frac{37.5}{100}=0.375$ From Carnot equation of efficiency, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $0.375=1-\frac{300}{\mathrm{~T}_{1}}$ $\frac{300}{\mathrm{~T}_{1}}=1-0.375$ $\frac{300}{\mathrm{~T}_{1}}=0.625$ $\mathrm{~T}_{1}=\frac{300}{0.625}=480 \mathrm{~K}=480-273=207^{\circ} \mathrm{C}$
AP EAMCET-05.10.2021
Thermodynamics
148643
A domestic refrigerator is loaded with food and the door closed. During a certain period the machine consumes $1 \mathrm{kWH}$ of energy and the internal energy of the system drops by $5000 \mathrm{~kJ}$. The quantity of heat transferred from the system to surrounding will be
1 $-8.6 \mathrm{MJ}$
2 $86 \mathrm{MJ}$
3 $-86 \mathrm{MJ}$
4 $-8.6 \mathrm{KJ}$
Explanation:
A Given, Power consumed regarded as work consumed which is taken as negative by sign convention $\therefore \mathrm{W}=-1 \mathrm{kWH}=-1 \times 3600 \mathrm{~kJ}=-3600 \mathrm{~kJ}$ $\Delta \mathrm{U}=-5000 \mathrm{~kJ}$ Applying first law of thermodynamics, $\mathrm{Q}-\mathrm{W}=\Delta \mathrm{U}$ $\mathrm{Q} -(-3600)=-5000$ $\therefore \mathrm{Q} =-5000-3600$ $\mathrm{Q} =-8600 \mathrm{~kJ}$ $\mathrm{Q} =-8.6 \mathrm{MJ}$
SRMJEEE-2019
Thermodynamics
148645
Inside the engine of an automobile, the cylinder compresses the air from approximately standard temperature and pressure to onetwentieth of the original volume and a pressure of $50 \mathrm{~atm}$. What is the temperature of the compressed air?
1 $500 \mathrm{~K}$
2 $682 \mathrm{~K}$
3 $550 \mathrm{~K}$
4 $1000 \mathrm{~K}$
5 $200 \mathrm{~K}$
Explanation:
B Given, $\mathrm{P}_{1}=1 \mathrm{~atm}, \mathrm{P}_{2}=50 \mathrm{~atm}, \mathrm{~V}_{1}=\mathrm{V}$, $\mathrm{V}_{2}=\frac{\mathrm{V}}{20} \& \mathrm{~T}_{1}=273 \mathrm{~K}, \mathrm{~T}_{2}=\text { ? }$ From ideal gas equation- $\mathrm{PV}=\mathrm{n} \text { RT }$ Taking ratio for two condition, $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{1 \times \mathrm{V}}{50 \times \frac{\mathrm{V}}{20}}=\frac{273}{\mathrm{~T}_{2}} \Rightarrow \frac{20}{50}=\frac{273}{\mathrm{~T}_{2}} \Rightarrow \mathrm{T}_{2}=\frac{273 \times 5}{2}$ $\mathrm{~T}_{2}=682 \mathrm{~K}$
Kerala CEE-2019
Thermodynamics
148646
An ideal mono atomic gas is taken round the cycle ABCDA as shown in figure. The work done during the cycle is :
1 PV
2 $2 \mathrm{PV}$
3 $\frac{P V}{2}$
4 zero
Explanation:
A Work done in a process is equal to the area under the curve $\mathrm{P}-\mathrm{V}$ diagram Work done during the cycle $=$ area of ABCDA $\mathrm{W} =\mathrm{BA} \times \mathrm{DA}$ $=(2 \mathrm{P}-\mathrm{P}) \times(2 \mathrm{~V}-\mathrm{V})$ $=\mathrm{PV}$
148639
A Carnot engine whose low-temperature reservoir is at $27^{\circ} \mathrm{C}$, has an efficiency $37.5 \%$. The high temperature reservoir is at
1 $480{ }^{\circ} \mathrm{C}$
2 $327^{\circ} \mathrm{C}$
3 $307^{\circ} \mathrm{C}$
4 $207^{\circ} \mathrm{C}$
Explanation:
D Given, $\mathrm{T}_{2}=27+273=300 \mathrm{~K}$ and efficiency $(\eta)=37.5 \%=\frac{37.5}{100}=0.375$ From Carnot equation of efficiency, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $0.375=1-\frac{300}{\mathrm{~T}_{1}}$ $\frac{300}{\mathrm{~T}_{1}}=1-0.375$ $\frac{300}{\mathrm{~T}_{1}}=0.625$ $\mathrm{~T}_{1}=\frac{300}{0.625}=480 \mathrm{~K}=480-273=207^{\circ} \mathrm{C}$
AP EAMCET-05.10.2021
Thermodynamics
148643
A domestic refrigerator is loaded with food and the door closed. During a certain period the machine consumes $1 \mathrm{kWH}$ of energy and the internal energy of the system drops by $5000 \mathrm{~kJ}$. The quantity of heat transferred from the system to surrounding will be
1 $-8.6 \mathrm{MJ}$
2 $86 \mathrm{MJ}$
3 $-86 \mathrm{MJ}$
4 $-8.6 \mathrm{KJ}$
Explanation:
A Given, Power consumed regarded as work consumed which is taken as negative by sign convention $\therefore \mathrm{W}=-1 \mathrm{kWH}=-1 \times 3600 \mathrm{~kJ}=-3600 \mathrm{~kJ}$ $\Delta \mathrm{U}=-5000 \mathrm{~kJ}$ Applying first law of thermodynamics, $\mathrm{Q}-\mathrm{W}=\Delta \mathrm{U}$ $\mathrm{Q} -(-3600)=-5000$ $\therefore \mathrm{Q} =-5000-3600$ $\mathrm{Q} =-8600 \mathrm{~kJ}$ $\mathrm{Q} =-8.6 \mathrm{MJ}$
SRMJEEE-2019
Thermodynamics
148645
Inside the engine of an automobile, the cylinder compresses the air from approximately standard temperature and pressure to onetwentieth of the original volume and a pressure of $50 \mathrm{~atm}$. What is the temperature of the compressed air?
1 $500 \mathrm{~K}$
2 $682 \mathrm{~K}$
3 $550 \mathrm{~K}$
4 $1000 \mathrm{~K}$
5 $200 \mathrm{~K}$
Explanation:
B Given, $\mathrm{P}_{1}=1 \mathrm{~atm}, \mathrm{P}_{2}=50 \mathrm{~atm}, \mathrm{~V}_{1}=\mathrm{V}$, $\mathrm{V}_{2}=\frac{\mathrm{V}}{20} \& \mathrm{~T}_{1}=273 \mathrm{~K}, \mathrm{~T}_{2}=\text { ? }$ From ideal gas equation- $\mathrm{PV}=\mathrm{n} \text { RT }$ Taking ratio for two condition, $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{1 \times \mathrm{V}}{50 \times \frac{\mathrm{V}}{20}}=\frac{273}{\mathrm{~T}_{2}} \Rightarrow \frac{20}{50}=\frac{273}{\mathrm{~T}_{2}} \Rightarrow \mathrm{T}_{2}=\frac{273 \times 5}{2}$ $\mathrm{~T}_{2}=682 \mathrm{~K}$
Kerala CEE-2019
Thermodynamics
148646
An ideal mono atomic gas is taken round the cycle ABCDA as shown in figure. The work done during the cycle is :
1 PV
2 $2 \mathrm{PV}$
3 $\frac{P V}{2}$
4 zero
Explanation:
A Work done in a process is equal to the area under the curve $\mathrm{P}-\mathrm{V}$ diagram Work done during the cycle $=$ area of ABCDA $\mathrm{W} =\mathrm{BA} \times \mathrm{DA}$ $=(2 \mathrm{P}-\mathrm{P}) \times(2 \mathrm{~V}-\mathrm{V})$ $=\mathrm{PV}$
148639
A Carnot engine whose low-temperature reservoir is at $27^{\circ} \mathrm{C}$, has an efficiency $37.5 \%$. The high temperature reservoir is at
1 $480{ }^{\circ} \mathrm{C}$
2 $327^{\circ} \mathrm{C}$
3 $307^{\circ} \mathrm{C}$
4 $207^{\circ} \mathrm{C}$
Explanation:
D Given, $\mathrm{T}_{2}=27+273=300 \mathrm{~K}$ and efficiency $(\eta)=37.5 \%=\frac{37.5}{100}=0.375$ From Carnot equation of efficiency, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $0.375=1-\frac{300}{\mathrm{~T}_{1}}$ $\frac{300}{\mathrm{~T}_{1}}=1-0.375$ $\frac{300}{\mathrm{~T}_{1}}=0.625$ $\mathrm{~T}_{1}=\frac{300}{0.625}=480 \mathrm{~K}=480-273=207^{\circ} \mathrm{C}$
AP EAMCET-05.10.2021
Thermodynamics
148643
A domestic refrigerator is loaded with food and the door closed. During a certain period the machine consumes $1 \mathrm{kWH}$ of energy and the internal energy of the system drops by $5000 \mathrm{~kJ}$. The quantity of heat transferred from the system to surrounding will be
1 $-8.6 \mathrm{MJ}$
2 $86 \mathrm{MJ}$
3 $-86 \mathrm{MJ}$
4 $-8.6 \mathrm{KJ}$
Explanation:
A Given, Power consumed regarded as work consumed which is taken as negative by sign convention $\therefore \mathrm{W}=-1 \mathrm{kWH}=-1 \times 3600 \mathrm{~kJ}=-3600 \mathrm{~kJ}$ $\Delta \mathrm{U}=-5000 \mathrm{~kJ}$ Applying first law of thermodynamics, $\mathrm{Q}-\mathrm{W}=\Delta \mathrm{U}$ $\mathrm{Q} -(-3600)=-5000$ $\therefore \mathrm{Q} =-5000-3600$ $\mathrm{Q} =-8600 \mathrm{~kJ}$ $\mathrm{Q} =-8.6 \mathrm{MJ}$
SRMJEEE-2019
Thermodynamics
148645
Inside the engine of an automobile, the cylinder compresses the air from approximately standard temperature and pressure to onetwentieth of the original volume and a pressure of $50 \mathrm{~atm}$. What is the temperature of the compressed air?
1 $500 \mathrm{~K}$
2 $682 \mathrm{~K}$
3 $550 \mathrm{~K}$
4 $1000 \mathrm{~K}$
5 $200 \mathrm{~K}$
Explanation:
B Given, $\mathrm{P}_{1}=1 \mathrm{~atm}, \mathrm{P}_{2}=50 \mathrm{~atm}, \mathrm{~V}_{1}=\mathrm{V}$, $\mathrm{V}_{2}=\frac{\mathrm{V}}{20} \& \mathrm{~T}_{1}=273 \mathrm{~K}, \mathrm{~T}_{2}=\text { ? }$ From ideal gas equation- $\mathrm{PV}=\mathrm{n} \text { RT }$ Taking ratio for two condition, $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{1 \times \mathrm{V}}{50 \times \frac{\mathrm{V}}{20}}=\frac{273}{\mathrm{~T}_{2}} \Rightarrow \frac{20}{50}=\frac{273}{\mathrm{~T}_{2}} \Rightarrow \mathrm{T}_{2}=\frac{273 \times 5}{2}$ $\mathrm{~T}_{2}=682 \mathrm{~K}$
Kerala CEE-2019
Thermodynamics
148646
An ideal mono atomic gas is taken round the cycle ABCDA as shown in figure. The work done during the cycle is :
1 PV
2 $2 \mathrm{PV}$
3 $\frac{P V}{2}$
4 zero
Explanation:
A Work done in a process is equal to the area under the curve $\mathrm{P}-\mathrm{V}$ diagram Work done during the cycle $=$ area of ABCDA $\mathrm{W} =\mathrm{BA} \times \mathrm{DA}$ $=(2 \mathrm{P}-\mathrm{P}) \times(2 \mathrm{~V}-\mathrm{V})$ $=\mathrm{PV}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
148639
A Carnot engine whose low-temperature reservoir is at $27^{\circ} \mathrm{C}$, has an efficiency $37.5 \%$. The high temperature reservoir is at
1 $480{ }^{\circ} \mathrm{C}$
2 $327^{\circ} \mathrm{C}$
3 $307^{\circ} \mathrm{C}$
4 $207^{\circ} \mathrm{C}$
Explanation:
D Given, $\mathrm{T}_{2}=27+273=300 \mathrm{~K}$ and efficiency $(\eta)=37.5 \%=\frac{37.5}{100}=0.375$ From Carnot equation of efficiency, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $0.375=1-\frac{300}{\mathrm{~T}_{1}}$ $\frac{300}{\mathrm{~T}_{1}}=1-0.375$ $\frac{300}{\mathrm{~T}_{1}}=0.625$ $\mathrm{~T}_{1}=\frac{300}{0.625}=480 \mathrm{~K}=480-273=207^{\circ} \mathrm{C}$
AP EAMCET-05.10.2021
Thermodynamics
148643
A domestic refrigerator is loaded with food and the door closed. During a certain period the machine consumes $1 \mathrm{kWH}$ of energy and the internal energy of the system drops by $5000 \mathrm{~kJ}$. The quantity of heat transferred from the system to surrounding will be
1 $-8.6 \mathrm{MJ}$
2 $86 \mathrm{MJ}$
3 $-86 \mathrm{MJ}$
4 $-8.6 \mathrm{KJ}$
Explanation:
A Given, Power consumed regarded as work consumed which is taken as negative by sign convention $\therefore \mathrm{W}=-1 \mathrm{kWH}=-1 \times 3600 \mathrm{~kJ}=-3600 \mathrm{~kJ}$ $\Delta \mathrm{U}=-5000 \mathrm{~kJ}$ Applying first law of thermodynamics, $\mathrm{Q}-\mathrm{W}=\Delta \mathrm{U}$ $\mathrm{Q} -(-3600)=-5000$ $\therefore \mathrm{Q} =-5000-3600$ $\mathrm{Q} =-8600 \mathrm{~kJ}$ $\mathrm{Q} =-8.6 \mathrm{MJ}$
SRMJEEE-2019
Thermodynamics
148645
Inside the engine of an automobile, the cylinder compresses the air from approximately standard temperature and pressure to onetwentieth of the original volume and a pressure of $50 \mathrm{~atm}$. What is the temperature of the compressed air?
1 $500 \mathrm{~K}$
2 $682 \mathrm{~K}$
3 $550 \mathrm{~K}$
4 $1000 \mathrm{~K}$
5 $200 \mathrm{~K}$
Explanation:
B Given, $\mathrm{P}_{1}=1 \mathrm{~atm}, \mathrm{P}_{2}=50 \mathrm{~atm}, \mathrm{~V}_{1}=\mathrm{V}$, $\mathrm{V}_{2}=\frac{\mathrm{V}}{20} \& \mathrm{~T}_{1}=273 \mathrm{~K}, \mathrm{~T}_{2}=\text { ? }$ From ideal gas equation- $\mathrm{PV}=\mathrm{n} \text { RT }$ Taking ratio for two condition, $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{1 \times \mathrm{V}}{50 \times \frac{\mathrm{V}}{20}}=\frac{273}{\mathrm{~T}_{2}} \Rightarrow \frac{20}{50}=\frac{273}{\mathrm{~T}_{2}} \Rightarrow \mathrm{T}_{2}=\frac{273 \times 5}{2}$ $\mathrm{~T}_{2}=682 \mathrm{~K}$
Kerala CEE-2019
Thermodynamics
148646
An ideal mono atomic gas is taken round the cycle ABCDA as shown in figure. The work done during the cycle is :
1 PV
2 $2 \mathrm{PV}$
3 $\frac{P V}{2}$
4 zero
Explanation:
A Work done in a process is equal to the area under the curve $\mathrm{P}-\mathrm{V}$ diagram Work done during the cycle $=$ area of ABCDA $\mathrm{W} =\mathrm{BA} \times \mathrm{DA}$ $=(2 \mathrm{P}-\mathrm{P}) \times(2 \mathrm{~V}-\mathrm{V})$ $=\mathrm{PV}$
