What is the resulting change in the internal energy of the system when 50 calories are added to
the system and the system does work of 30 calories on the surroundings?

1. 20 cal                             
2. 50 cal
3. 40 cal                             
4. 30 cal

A gaseous system changes from state A(P₁, V₁, T₁) to B(P₂,V₂,T₂), B to C(P₃, V₃, T₃) and finally from C to A. The whole process may be called:

1. reversible process                           

2. cyclic process

3. isobaric process                               

4. spontaneous process

Which statements are correct?

1. $2.303<mspace\; width="1em"></mspace>\log \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{∆{\mathrm{H}}_{\mathrm{vap}.}}{\mathrm{R}}\frac{\left[{\mathrm{T}}_2-{\mathrm{T}}_1\right]}{{\mathrm{T}}_1{\mathrm{T}}_2}$ is called Clausius-Clapeyron equation

2. $\frac{∆{\mathrm{H}}_{\mathrm{vap}.}}{\mathrm{Boiling}<mspace\; width="1em"></mspace>\mathrm{Point}}=88<mspace\; width="1em"></mspace>J<mspace\; width="1em"></mspace>mo{l}^{-1}{K}^{-1}$ is called Trouton's rule

3. Entropy is a measure of unavailable energy, i.e.,

     unavailable energy = entropy x temperature

4. All of the above

One mole of ice is converted to water at 273 K. The entropies of H₂O(s) and H₂O(l) are 38.20 J mol⁻¹K⁻¹ and 60.01 J mol⁻¹K⁻¹, respectively. What is the enthalpy change for this conversion?

1. 59.54 J mol^-1                               

2. 5954 J mol^-1

3. 595.4 J mol^-1                             

4. 320.6 J mol^-1

$∆\mathrm{S}°$ will be highest for the reaction:

1. Ca(s) + 1/2 O₂(g) $\to$CaO(s)

2. CaCO₃(s)$\to$CaO(s) + CO₂(g)

3. C(s) + O₂(g)$\to$CO₂(g)

4. N₂(g) + O₂(g)$\to$2NO(g)

Change in entropy is negative for:

1. Bromine (l)$\to$Bromine(g)

2. C(s) + H₂O(g) $\to$CO(g) + H₂(g)

3. N₂(g,10 atm)$\to$N₂(g,1 atm) 

4. Fe ( 1mol, 400 K) $\to$ Fe( 1mol, 300 K)

The internal energy change when a system goes from state A to B is 40 kJ/mol. If the system goes from A to B by a reversible path and returns to state A by an irreversible path. What would be the change in internal energy?

1. 40 kJ                                   

2. >40 kJ

3. <40 kJ                                 

4. Zero

The mathematical form of the first law of thermodynamics when heat (q) is supplied and W is work done by the system (-ve) is:

1. $∆$U=q+W                       

2. $∆$U=q-W

3. $∆$U=-q+W                       

4. $∆$U= -q-W

For the process:

H₂O(l)[1 bar, 373 K] $\rightleftharpoons$H₂O(g)[1 bar, 373 K] the correct set of thermodynamic parameters are:

1. $∆\mathrm{G}=0;<mspace\; width="1em"></mspace>∆\mathrm{S}=+\mathrm{ve}$

2. $∆\mathrm{G}=0;<mspace\; width="1em"></mspace>∆\mathrm{S}=-\mathrm{ve}$

3. $∆\mathrm{G}=+\mathrm{ve};<mspace\; width="1em"></mspace>∆\mathrm{S}=0$

4. $∆\mathrm{G}=-\mathrm{ve};<mspace\; width="1em"></mspace>∆\mathrm{S}=+\mathrm{ve}$