Heat of combustion ∆Hº for C(s), H2(g) and CH4(g) are – 94, – 68 and – 213 Kcal/mol. ∆Hº for C(s) + 2H2(g) → CH4 (g) is:

1. – 17 Kcal 2. – 111 Kcal
3. – 170 Kcal 4. – 85 Kcal

Subtopic:  Thermochemistry |
 82%
From NCERT
AIPMT - 2002
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When 1 mol gas is heated at constant volume, the temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. The correct statement among the following is:
1.  q = w = 500 J, ∆U = 0
2.  q = ∆U = 500 J, w = 0
3.  q = w = 500 J, ∆U = 0
4.  ∆U = 0, q = w = – 500 J

Subtopic:  First Law of Thermodynamics |
 86%
From NCERT
AIPMT - 2001
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The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm–3, respectively. If the standard free energy difference (∆Gº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is:

1. 11.08×108 Pa

2. 9.92×107 Pa

3. 9.92×106 Pa

4. 11.08×105 Pa

Subtopic:  Gibbs Energy Change |
AIPMT - 2003
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What is the entropy change (in JK–1 mol–1) when one mole of ice is converted into water at 0 ºC? (The enthalpy change for the conversion of ice to liquid water is 6.0 KJ mol–1 at 0 ºC)

1. 20.13 2. 2.013
3. 2.198 4. 21.98
Subtopic:  Spontaneity & Entropy |
 73%
From NCERT
AIPMT - 2003
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The formation of a solution from two components can be considered as:

(i) Pure solvent → separated solvent molecules, ∆H1
(ii) Pure solute → separated solute molecules, ∆H2
(iii) Separated solvent and solute molecules → solution, ∆H3

The solution so formed will be ideal if:
1. ∆HSoln = ∆H1 + ∆H2 + ∆H3

2. ∆HSoln = ∆H1 + ∆H2 – ∆H3

3. ∆HSoln = ∆H1 – ∆H2 – ∆H3

4. ∆HSoln = ∆H3 – ∆H1 – ∆H2

Subtopic:  Thermochemistry |
From NCERT
AIPMT - 2003
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NEET 2023 - Target Batch - Aryan Raj Singh
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The molar heat capacity of water at constant pressure, C, is 75 JK–1 mol–1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of the water is:

1. 1.2 K 2. 2.4 K
3. 4.8 K 4. 6.6 K
Subtopic:  First Law of Thermodynamics |
 76%
From NCERT
AIPMT - 2003
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12N2 (g) + 12O2 (g)  NO(g) ;
rH° = 90 kJ mol-1
NO(g) + 12O2(g)  NO2(g); 
rH° = -74 kJ mol-1

The thermodynamic stability of NO(g) based on the above data is:

1. Less than NO2(g) 

2. More than NO2(g)

3. Equal to NO2(g)

4. Insufficient data

Subtopic:  Enthalpy & Internal energy |
 71%
From NCERT
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The entropy change in the surroundings when 1.00 mol of H2O(l) is formed under standard conditions is-

fHθ = –286 kJ mol–1 

1. 952.5 J mol-1

2. 979.7 J mol-1

3. 949.7 J mol-1

4. 959.7 J mol-1

Subtopic:  Spontaneity & Entropy |
 50%
From NCERT
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For the graph given below, it can be concluded that work done during the process shown will be-

1. Zero 2. Negative
3. Positive 4. Cannot be determined
Subtopic:  First Law of Thermodynamics |
 65%
From NCERT
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Consider the following graph.

  

The work done shown by the above-mentioned graph is-

1. Positive 2. Negative
3. Zero 4. Cannot be determined
Subtopic:  First Law of Thermodynamics |
 70%
From NCERT
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