272677
Calculate the work done when a gas is compressed by an average pressure of $0.50 \mathrm{~atm}$ so as to decrease its volume from $400 \mathrm{~cm}^3$ to $200 \mathrm{~cm}^3$.
272678
A system is provided with $50 \mathrm{~J}$ of heat and the work done on the system is $10 \mathrm{~J}$. What is the change in internal energy of the system in joules?
1 60
2 40
3 50
4 10
Explanation:
Given, $\mathrm{q}=50 \mathrm{~J}$ $\mathrm{~W}=10 \mathrm{~J}$ We know that, The law of conservation of energy that the total is constant, energy con be transformed from one from to another, but can be neither created or destroyed. $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\Delta \mathrm{U}=50+10=60 \mathrm{~J} .$
AP-EAMCET- (Engg.) - 2010
Thermodynamics
272679
Calculate the maximum work done in expanding $16 \mathrm{~g}$ of oxygen at $300 \mathrm{~K}$ and occupying a volume of $5 \mathrm{dm}^3$ isothermally until volume becomes $25 \mathrm{dm}^3$.
1 $-2.01 \times 10^3 \mathrm{~J}$
2 $+2.01 \times 10^3 \mathrm{~J}$
3 $2.01 \mathrm{~J}$
4 $2.01 \times 10^3 \mathrm{~kJ}$
Explanation:
Maximum work is obtained in isothermal reversible process. The expansion for the maximum work is as Follow, $\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_2}{\mathrm{~V}_1}, \mathrm{~W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2}$ The number of moles of oxygen (n) $=$ $\frac{\text { Weight }}{\text { Molecular Weight }}=\frac{16}{32}=0.5$ The temperature is $T=300 \mathrm{~K}$ The initial volume is $\mathrm{V}_1=5 \mathrm{dm}^3$ The final volume is $V_2=25 \mathrm{dm}^3$ Substituting values in the above reaction we get, $\mathrm{W}=-2.303 \times 0.5 \times 8.314 \times 300 \times \log \frac{25}{5}=-2.01 \times 10^3 \mathrm{~J}$
AMU-2017
Thermodynamics
272684
When 1 mole gas is heated at constant volume, temperature is raised from $298 \mathrm{~K}$ to $308 \mathrm{~K}$. Heat supplied to the gas is 500J. Then which statement is correct?
272677
Calculate the work done when a gas is compressed by an average pressure of $0.50 \mathrm{~atm}$ so as to decrease its volume from $400 \mathrm{~cm}^3$ to $200 \mathrm{~cm}^3$.
272678
A system is provided with $50 \mathrm{~J}$ of heat and the work done on the system is $10 \mathrm{~J}$. What is the change in internal energy of the system in joules?
1 60
2 40
3 50
4 10
Explanation:
Given, $\mathrm{q}=50 \mathrm{~J}$ $\mathrm{~W}=10 \mathrm{~J}$ We know that, The law of conservation of energy that the total is constant, energy con be transformed from one from to another, but can be neither created or destroyed. $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\Delta \mathrm{U}=50+10=60 \mathrm{~J} .$
AP-EAMCET- (Engg.) - 2010
Thermodynamics
272679
Calculate the maximum work done in expanding $16 \mathrm{~g}$ of oxygen at $300 \mathrm{~K}$ and occupying a volume of $5 \mathrm{dm}^3$ isothermally until volume becomes $25 \mathrm{dm}^3$.
1 $-2.01 \times 10^3 \mathrm{~J}$
2 $+2.01 \times 10^3 \mathrm{~J}$
3 $2.01 \mathrm{~J}$
4 $2.01 \times 10^3 \mathrm{~kJ}$
Explanation:
Maximum work is obtained in isothermal reversible process. The expansion for the maximum work is as Follow, $\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_2}{\mathrm{~V}_1}, \mathrm{~W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2}$ The number of moles of oxygen (n) $=$ $\frac{\text { Weight }}{\text { Molecular Weight }}=\frac{16}{32}=0.5$ The temperature is $T=300 \mathrm{~K}$ The initial volume is $\mathrm{V}_1=5 \mathrm{dm}^3$ The final volume is $V_2=25 \mathrm{dm}^3$ Substituting values in the above reaction we get, $\mathrm{W}=-2.303 \times 0.5 \times 8.314 \times 300 \times \log \frac{25}{5}=-2.01 \times 10^3 \mathrm{~J}$
AMU-2017
Thermodynamics
272684
When 1 mole gas is heated at constant volume, temperature is raised from $298 \mathrm{~K}$ to $308 \mathrm{~K}$. Heat supplied to the gas is 500J. Then which statement is correct?
272677
Calculate the work done when a gas is compressed by an average pressure of $0.50 \mathrm{~atm}$ so as to decrease its volume from $400 \mathrm{~cm}^3$ to $200 \mathrm{~cm}^3$.
272678
A system is provided with $50 \mathrm{~J}$ of heat and the work done on the system is $10 \mathrm{~J}$. What is the change in internal energy of the system in joules?
1 60
2 40
3 50
4 10
Explanation:
Given, $\mathrm{q}=50 \mathrm{~J}$ $\mathrm{~W}=10 \mathrm{~J}$ We know that, The law of conservation of energy that the total is constant, energy con be transformed from one from to another, but can be neither created or destroyed. $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\Delta \mathrm{U}=50+10=60 \mathrm{~J} .$
AP-EAMCET- (Engg.) - 2010
Thermodynamics
272679
Calculate the maximum work done in expanding $16 \mathrm{~g}$ of oxygen at $300 \mathrm{~K}$ and occupying a volume of $5 \mathrm{dm}^3$ isothermally until volume becomes $25 \mathrm{dm}^3$.
1 $-2.01 \times 10^3 \mathrm{~J}$
2 $+2.01 \times 10^3 \mathrm{~J}$
3 $2.01 \mathrm{~J}$
4 $2.01 \times 10^3 \mathrm{~kJ}$
Explanation:
Maximum work is obtained in isothermal reversible process. The expansion for the maximum work is as Follow, $\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_2}{\mathrm{~V}_1}, \mathrm{~W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2}$ The number of moles of oxygen (n) $=$ $\frac{\text { Weight }}{\text { Molecular Weight }}=\frac{16}{32}=0.5$ The temperature is $T=300 \mathrm{~K}$ The initial volume is $\mathrm{V}_1=5 \mathrm{dm}^3$ The final volume is $V_2=25 \mathrm{dm}^3$ Substituting values in the above reaction we get, $\mathrm{W}=-2.303 \times 0.5 \times 8.314 \times 300 \times \log \frac{25}{5}=-2.01 \times 10^3 \mathrm{~J}$
AMU-2017
Thermodynamics
272684
When 1 mole gas is heated at constant volume, temperature is raised from $298 \mathrm{~K}$ to $308 \mathrm{~K}$. Heat supplied to the gas is 500J. Then which statement is correct?
NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
272677
Calculate the work done when a gas is compressed by an average pressure of $0.50 \mathrm{~atm}$ so as to decrease its volume from $400 \mathrm{~cm}^3$ to $200 \mathrm{~cm}^3$.
272678
A system is provided with $50 \mathrm{~J}$ of heat and the work done on the system is $10 \mathrm{~J}$. What is the change in internal energy of the system in joules?
1 60
2 40
3 50
4 10
Explanation:
Given, $\mathrm{q}=50 \mathrm{~J}$ $\mathrm{~W}=10 \mathrm{~J}$ We know that, The law of conservation of energy that the total is constant, energy con be transformed from one from to another, but can be neither created or destroyed. $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\Delta \mathrm{U}=50+10=60 \mathrm{~J} .$
AP-EAMCET- (Engg.) - 2010
Thermodynamics
272679
Calculate the maximum work done in expanding $16 \mathrm{~g}$ of oxygen at $300 \mathrm{~K}$ and occupying a volume of $5 \mathrm{dm}^3$ isothermally until volume becomes $25 \mathrm{dm}^3$.
1 $-2.01 \times 10^3 \mathrm{~J}$
2 $+2.01 \times 10^3 \mathrm{~J}$
3 $2.01 \mathrm{~J}$
4 $2.01 \times 10^3 \mathrm{~kJ}$
Explanation:
Maximum work is obtained in isothermal reversible process. The expansion for the maximum work is as Follow, $\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_2}{\mathrm{~V}_1}, \mathrm{~W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2}$ The number of moles of oxygen (n) $=$ $\frac{\text { Weight }}{\text { Molecular Weight }}=\frac{16}{32}=0.5$ The temperature is $T=300 \mathrm{~K}$ The initial volume is $\mathrm{V}_1=5 \mathrm{dm}^3$ The final volume is $V_2=25 \mathrm{dm}^3$ Substituting values in the above reaction we get, $\mathrm{W}=-2.303 \times 0.5 \times 8.314 \times 300 \times \log \frac{25}{5}=-2.01 \times 10^3 \mathrm{~J}$
AMU-2017
Thermodynamics
272684
When 1 mole gas is heated at constant volume, temperature is raised from $298 \mathrm{~K}$ to $308 \mathrm{~K}$. Heat supplied to the gas is 500J. Then which statement is correct?
