Test: Ideal Gas Equation & Dalton's Law of Partial Pressure (Old NCERT) - JEE MCQ

P₁V₁/T₁ = P₂V₂/T₂
Let the original volume be 3 volume units . The final volume will be 1/3 of 3 which is 1 volume unit
Temperature must be in K - 0^οC = 273K and 546^οC = 819 K
We want to solve for P1
P₁*1/819 = 5*3/273
P₁ = 819*5*3/273
P₁ = 45

By applying ideal gas eqn. PV = nRT
Further solving it we get an eq. 
w = PVM/RT 
Putting values, we get 64g.
Hence, the correct answer is Option A.

Applying pV = nRT in state 1, we have n₁ = 0.110 
Applying pV = nRT in state 1, we have n₂ = 0.0184
Mole fraction(this fraction is present) = 0.0184/0.110 = 0.16727
So, fraction escaped = (1-0.16727)*100 = 83.33%

The correct answer is Option C.

The average molar mass of the vapour above solid NH₄Cl = 26.5 mol^-1
Evaporation product of NH4Cl --->
  NH₄Cl (s) ----> NH₃(g) + HCl(g)
Let Mole fraction of NH₃ = X_L
       Mole fraction of HCl = X₂
∴ Average molar mass of NH₃ = 14 + 31 = 17
=> 26.5 = m₁x₁ + m2x2
=> 26.5 = 17x1 + 36.5x₂    ---(i)
And we know;
        x₁ + x₂ = 1
      From (i) and (ii) 
x1 = 0.5 and x₂ = 0.5

Given pressure(P) of gas A= 2bar Total pressure = 3bar then pressure(P) of gas B = 3-2 =1bar
no. of mole (n) = given mass(m) / molar mass(M)
Gas A
PV = nRT
2×V =( mₐ/Mₐ) RT ------------- (1)
Gas B
PV = nRT
1V = (mb/Mb)RT ---------------- (2)
On dividing eqn 1 by eqn 2
2V/1V = (1/Mₐ) ÷(2/Mb)
2/1 = (Mb/Mₐ) × 1/2
(M_B/Mₐ) = (2/1)×(2/1)
M_B/Mₐ = 4/1
M_B = 4Mₐ

Correct answer is option B
Avagadro′s Law says:
Equal volume of two gases at same pressure and temperature conditions have equal number of molecules.
∴1 O₂ molecule consumes 2 H₂ molecules.    
∴Since after the reaction only H₂ is left,  we know all of O₂ is consumed.
∴This implies what all consumed 1/3 of it was O₂.
Hence, O₂ in the mix was at a partial pressure of (1 - 0.35)* 1/3 = 0.22
And rest was H₂ = (1 - 0.22) = 0.78
Hence oxygen- hydrogen composition in the original mix was:0.22 bar O2 and 0.78 bar H₂.

Applying p1V1 + p2V2 + p3V3 = pnet × Vnet
(0.792)×4 + (1.23)×3 + (2.51)×5 = pnet × 12
On solving, we get p = 1.617 = 1.62 atm

Correct option is not provided. 
    2H2     +    O2 ------>   2H2O
      3         1                0
(-)   2         1                2 mole H₂O formed
___________________________
    1               0              Total mole = 1+2 = 3

PV = nRT / 100
  n = PV / RT
  P = nRT/ V
      = [4 * 0.083 * (125 + 273)] / 125.3
Pressure = 1.0545 bar after water formed is 2 mole.
P = [3 * 0.083 * 398] /125.3
   = 0.7909 bar
   = 0.791 bar (appx)
Hence, the answer is 1.0545 bar, 0.7909 bar. 
 

n_NO2 + 2n_N2O4= 0.330 mol
Subtracting n_NO2+ n_N2O4 = 0.204 mol
From n_NO2+ 2n_N2O4 = 0.330 mol
n_NO2 = 0.126 mol
n_N2O4 = 0.078 mol
From these results, NO₂ has a mole fraction of 0.62and a partial pressure of 0.62(0.50 atm) = 0.31 atm;N2O4has a mole fraction of 0.38 and a partial pressure of 0.38(0.50 atm) = 0.19 atm

We know that
pV = nRT
A/Q , p and T are constant;
So  n₁/n₂ = V₁/V₂
(¼)/(0.25/4) = (10)3/r3
r3 = 750
r  ∼ 9

The correct answer is 2
The solution is as follows:
Δm = m_cold − m_hot     
 =[n_cold−n_hot]M  
   = PVMR[1T1− 1T2]
Solving we get:
T2−T1 = ΔT = 2

The empirical formula is = Fe(CO)₂
Carbon dioxide obtained:
Pressure = 44.9 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 44.9 / 760 = 0.0591 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15  
So,  
T₁ = (25 + 273.15) K = 298.15 K  
V = 1.50 L
Using ideal gas equation as:
PV=nRT
where,  
P is the pressure
V is the volume
n is the number of moles
T is the temperature  
R is Gas constant having value 
= 0.0821 L.atm/K.mol
Applying the equation as:
0.0591 atm × 1.50 L 
= n × 0.0821 L.atm/K.mol × 298.15 K  
⇒n = 0.0036 moles
1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,
Moles of C = 0.0036 moles
Molar mass of C atom 
= 12.0107 g/mol
Mass of C in molecule 
= 0.0036 x 12.0107 = 0.0432 g
Given that the compound only contains iron and carbon. So,
Mass of Fe in the sample 
= Total mass - Mass of C
Mass of the sample = 0.142 g
Mass of Fe in sample 
= 0.142 g - 0.0432 g = 0.0988 g  
Molar mass of Fe = 55.845 g/mol
Moles of Fe  
= 0.0988  / 55.845  = 0.0018 moles
Taking the simplest ratio for Fe and C as:
0.0018 : 0.0036
 = 1 : 2
 

Let nx mole of unknown compound be present in vapour form so its partial pressure is
0.68 = (nx/nx+0.1)×1 atm
nx+0.1/nx = 1/0.68 = 1.47
0.1/nx = 0.47
nx = 0.212
Applying ideal gas equation,
1×V = (0.212/0.1)×0.0821×273
V = 7