272712
The first law of thermodynamics for isothermal process is
1 $\mathrm{q}=-\mathrm{W}$
2 $\Delta \mathrm{U}=\mathrm{W}$
3 $\Delta \mathrm{U}=\mathrm{q}_{\mathrm{v}}$
4 $\Delta \mathrm{U}=-\mathrm{q}_{\mathrm{v}}$
Explanation:
According to the first law of thermodynamics $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ For an isothermal process, $\Delta \mathrm{U}=0$ Hence, $\mathrm{q}+\mathrm{W}=0$ or $q=-W$
MHT CET-2017
Thermodynamics
272719
$16 \mathrm{~g}$ of oxygen gas expands isothermally and reversible at $300 \mathrm{~K}$ from $10 \mathrm{dm}^3$ to $100 \mathrm{dm}^3$ the work done is (in $J$ )
1 zero
2 $-2875 \mathrm{~J}$
3 $+2875 \mathrm{~J}$
4 infinite
Explanation:
$\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_2}{\mathrm{~V}_1}$ As we know that, work done in an isothermal reversible expansion is given as - $W =-2.303 \times \frac{16}{32} \times 8.314 \times 300 \log \frac{100}{10}$ $=-2872 \mathrm{~J} \approx-2875 \mathrm{~J}$
MHT CET-2009
Thermodynamics
272725
Which of the following is the correct equation?
First law of thermodynamics, $\Delta \mathrm{U}=\Delta \mathrm{Q}+\Delta \mathrm{W}$ Where, $\Delta \mathrm{U}=$ Change in internal energy $\Delta Q=$ Heat absorbed by the system $\Delta \mathrm{W}=$ Workdone by the system
NEET-1996
Thermodynamics
272700
The correct statement regarding entropy is
1 at absolute zero temperature, entropy of a perfectly crystalline solid in zero
2 at absolute zero temperature, the entropy of a perfectly crystalline substance is positive
3 at absolute zero temperature, the entropy of all crystalline substances is zero
4 at $0^{\circ} \mathrm{C}$, the entropy of a perfect crystalline solid in zero
Explanation:
Entropy of perfectly crystalline solids will be zero at absolute zero. It is predicted in the third law of thermodynamics.
272712
The first law of thermodynamics for isothermal process is
1 $\mathrm{q}=-\mathrm{W}$
2 $\Delta \mathrm{U}=\mathrm{W}$
3 $\Delta \mathrm{U}=\mathrm{q}_{\mathrm{v}}$
4 $\Delta \mathrm{U}=-\mathrm{q}_{\mathrm{v}}$
Explanation:
According to the first law of thermodynamics $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ For an isothermal process, $\Delta \mathrm{U}=0$ Hence, $\mathrm{q}+\mathrm{W}=0$ or $q=-W$
MHT CET-2017
Thermodynamics
272719
$16 \mathrm{~g}$ of oxygen gas expands isothermally and reversible at $300 \mathrm{~K}$ from $10 \mathrm{dm}^3$ to $100 \mathrm{dm}^3$ the work done is (in $J$ )
1 zero
2 $-2875 \mathrm{~J}$
3 $+2875 \mathrm{~J}$
4 infinite
Explanation:
$\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_2}{\mathrm{~V}_1}$ As we know that, work done in an isothermal reversible expansion is given as - $W =-2.303 \times \frac{16}{32} \times 8.314 \times 300 \log \frac{100}{10}$ $=-2872 \mathrm{~J} \approx-2875 \mathrm{~J}$
MHT CET-2009
Thermodynamics
272725
Which of the following is the correct equation?
First law of thermodynamics, $\Delta \mathrm{U}=\Delta \mathrm{Q}+\Delta \mathrm{W}$ Where, $\Delta \mathrm{U}=$ Change in internal energy $\Delta Q=$ Heat absorbed by the system $\Delta \mathrm{W}=$ Workdone by the system
NEET-1996
Thermodynamics
272700
The correct statement regarding entropy is
1 at absolute zero temperature, entropy of a perfectly crystalline solid in zero
2 at absolute zero temperature, the entropy of a perfectly crystalline substance is positive
3 at absolute zero temperature, the entropy of all crystalline substances is zero
4 at $0^{\circ} \mathrm{C}$, the entropy of a perfect crystalline solid in zero
Explanation:
Entropy of perfectly crystalline solids will be zero at absolute zero. It is predicted in the third law of thermodynamics.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
272712
The first law of thermodynamics for isothermal process is
1 $\mathrm{q}=-\mathrm{W}$
2 $\Delta \mathrm{U}=\mathrm{W}$
3 $\Delta \mathrm{U}=\mathrm{q}_{\mathrm{v}}$
4 $\Delta \mathrm{U}=-\mathrm{q}_{\mathrm{v}}$
Explanation:
According to the first law of thermodynamics $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ For an isothermal process, $\Delta \mathrm{U}=0$ Hence, $\mathrm{q}+\mathrm{W}=0$ or $q=-W$
MHT CET-2017
Thermodynamics
272719
$16 \mathrm{~g}$ of oxygen gas expands isothermally and reversible at $300 \mathrm{~K}$ from $10 \mathrm{dm}^3$ to $100 \mathrm{dm}^3$ the work done is (in $J$ )
1 zero
2 $-2875 \mathrm{~J}$
3 $+2875 \mathrm{~J}$
4 infinite
Explanation:
$\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_2}{\mathrm{~V}_1}$ As we know that, work done in an isothermal reversible expansion is given as - $W =-2.303 \times \frac{16}{32} \times 8.314 \times 300 \log \frac{100}{10}$ $=-2872 \mathrm{~J} \approx-2875 \mathrm{~J}$
MHT CET-2009
Thermodynamics
272725
Which of the following is the correct equation?
First law of thermodynamics, $\Delta \mathrm{U}=\Delta \mathrm{Q}+\Delta \mathrm{W}$ Where, $\Delta \mathrm{U}=$ Change in internal energy $\Delta Q=$ Heat absorbed by the system $\Delta \mathrm{W}=$ Workdone by the system
NEET-1996
Thermodynamics
272700
The correct statement regarding entropy is
1 at absolute zero temperature, entropy of a perfectly crystalline solid in zero
2 at absolute zero temperature, the entropy of a perfectly crystalline substance is positive
3 at absolute zero temperature, the entropy of all crystalline substances is zero
4 at $0^{\circ} \mathrm{C}$, the entropy of a perfect crystalline solid in zero
Explanation:
Entropy of perfectly crystalline solids will be zero at absolute zero. It is predicted in the third law of thermodynamics.
272712
The first law of thermodynamics for isothermal process is
1 $\mathrm{q}=-\mathrm{W}$
2 $\Delta \mathrm{U}=\mathrm{W}$
3 $\Delta \mathrm{U}=\mathrm{q}_{\mathrm{v}}$
4 $\Delta \mathrm{U}=-\mathrm{q}_{\mathrm{v}}$
Explanation:
According to the first law of thermodynamics $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ For an isothermal process, $\Delta \mathrm{U}=0$ Hence, $\mathrm{q}+\mathrm{W}=0$ or $q=-W$
MHT CET-2017
Thermodynamics
272719
$16 \mathrm{~g}$ of oxygen gas expands isothermally and reversible at $300 \mathrm{~K}$ from $10 \mathrm{dm}^3$ to $100 \mathrm{dm}^3$ the work done is (in $J$ )
1 zero
2 $-2875 \mathrm{~J}$
3 $+2875 \mathrm{~J}$
4 infinite
Explanation:
$\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_2}{\mathrm{~V}_1}$ As we know that, work done in an isothermal reversible expansion is given as - $W =-2.303 \times \frac{16}{32} \times 8.314 \times 300 \log \frac{100}{10}$ $=-2872 \mathrm{~J} \approx-2875 \mathrm{~J}$
MHT CET-2009
Thermodynamics
272725
Which of the following is the correct equation?
First law of thermodynamics, $\Delta \mathrm{U}=\Delta \mathrm{Q}+\Delta \mathrm{W}$ Where, $\Delta \mathrm{U}=$ Change in internal energy $\Delta Q=$ Heat absorbed by the system $\Delta \mathrm{W}=$ Workdone by the system
NEET-1996
Thermodynamics
272700
The correct statement regarding entropy is
1 at absolute zero temperature, entropy of a perfectly crystalline solid in zero
2 at absolute zero temperature, the entropy of a perfectly crystalline substance is positive
3 at absolute zero temperature, the entropy of all crystalline substances is zero
4 at $0^{\circ} \mathrm{C}$, the entropy of a perfect crystalline solid in zero
Explanation:
Entropy of perfectly crystalline solids will be zero at absolute zero. It is predicted in the third law of thermodynamics.