Thermodynamics: Basic Concepts - Free MCQ Practice Test with solutions,

Work done = - (change in moles) RT For work done positive change in moles negative and vice versa We only cosider the moles that are in gass phase.

In this process, one molecule of hydrogen gas dissociated into two gaseous atoms i.e. the number of gaseous particles increase leading to  a more disordered state. 
.Breaking of bond requires energy

It is not mentioned that the process is isobaric. If we assume that process is isobaric because volume can not remain constant as number of gaseous particles are increased, then change in internal energy = 436 kJ mol^-1. The data of absorption gives us the value of q.

Let atomic weight of x = M_x
atomic weight of y = M_y 

we know, 
mole = weight /atomic weight 
a/c to question, 

mole of xy₂ = 0.1 
so, 
0.1 = 10g/( M_x +2M_y) 

M_x + 2M_y = 100g -------(1)

for x₃y₂ ; mole of x₃y₂ = 0.05 

0.05 = 9/( 3M_x + 2M_y ) 

3M_x + 2M_y = 9/0.05 = 9 × 20 = 180 g ---(2) 
solve eqns (1) and (2)

2M_x = 80 
M_x = 40g/mol 

and M_y = 30g/mole

Work done during isothermal irreversible process is given by the formula w = -p_ext (nrt/p₂-nrt/p₁)

Use ∆H=∆U+∆NgRT FOR 1. we need to find work done in constant volume that is ∆U which will be 0 as it is nCv∆T and since ∆T is zero so ∆U is zero .
For 2. we need to Find ∆H =∆U+∆NgRT Since ∆U is zero so∆H=∆NgRT 
R=2 T =300and ∆Ng is = -1

Intensive means that the property is mass independent. We can see that Boiling point, pH, EMF of cell and surface tension are intensive pont . 
While Extensive property  depends on mass and in given options Volume and entropy are mass dependents.

Correct answer: D.

A state function is a property whose value depends only on the current state of the system and not on the path taken to reach that state.

Internal energy (I), Entropy (II), Free energy (III) and Enthalpy (IV) are all state functions because their values are determined by the state variables (for example, temperature, pressure, volume, composition) and do not depend on the process path.

Work is a path function; its magnitude depends on the process by which the change is carried out. Therefore both irreversible expansion work (V) and reversible expansion work (VI) are path functions and are not state functions.

Because only I, II, III, IV are state functions, the correct choice is option D.

In endothermic reactions, the energy of reactants is less than that of the products. Potential energy diagram for endothermic reactions is,

Where Ea = activation energy of forwarding reaction

Ea' = activation energy of backwards reaction

ΔH = enthalpy of the reaction

From the above diagram,

Ea = Ea' + ΔH

Thus, Ea > ΔH

Applying ideal gas eqn. at 1
p×22.4 = 1×0.0821×298        (since vol. Is in L. R should haave constaant with lit in it.)
p = 1.09 atm. To convert it into bar, we multiply it by 1.01 or we have 1.03 bar
Similarly at 2 and 3, we get value of pressure at 2 and 3. 
The most closest answer is option d

i) CaC₂ + 2H₂O  →  C₂H₂(g) + Ca(OH)₂
ii) Mg₂C₃ + 4H₂O → CH_3-C≡CH(g) + 2Mg(OH)₂
iii) Al₄C₃ + 12H₂O → 3CH₄(g) + 4Al(OH)₃
Work done = -∆n_gRT 
Or W ∝ ∆n_g
Wecan see that in case iii) we have maximum no of moles in the product side(∆n_g = 3). So work done will be maximum in case iii). After that in i) and ii), we have the same number of moles on the product side(∆n_g = 1), so work done will be the same.
Therefore, option c is correct.

The correct answer is Option B.
 
H−N−H (three N-H) bonds
     ∣
    H
Thus, ΔE=3xKJmol^-1

∆U = nfRT/2
Since He is monoatomic, so f(degree of freedom) = 3
∆U = 3×1×2×2/2   (We have taken R = 2 cal K^-1 mol ^-1)
= 6 cal

We know that for a cyclic process the net change in internal energy is equal to zero and change in the internal energy does not depend on the path by which the final state is reached.

From ideal gas equation 
PV=nRT 
hence PV=W
W = nR∆T 
W = 1×2×1 = 2cal

∆T, ∆p, ∆V and ∆U = 0 (also they are state function)
∆U = q+w. So, q and w are not zero.
Hence we have 4 parameters equal to zero after the whole process.

For heat added at constant volume, Work doe = 0
∆U = _T1∫^T2 nC_v,mdT
Or ∆U = nC_v,m∆T
∆T = ∆U/nC_v,m = 100/1×33.33 = 3
So, the correct answer is 3

Isochoric change is one in which volume doesn't change(remains constant) From P₁ to P₂ , volume is constant(V₁) (also work done = 0)

Work done is the area enclosed by PV graph with volume axis.