Test: Basic Concepts of Thermodynamics & Applications (June 12) - NEET MCQ

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.

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

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

First 4 are fundamental examples of state functions as they are path independent. Since reversible expansion depends on the path followed by process, so it is a path function. However, irreversible expansion work is independent of the path and so, it is a state function.

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.

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.

∆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

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