Equal weights of methane and hydrogen are mixed in anempty container at 25ºC. The fraction of the total pressure exerted by hydrogen is :(a)21(b)98(c)91(d)17

¹⁄₂
⁸⁄₉
¹⁄₉
¹⁶⁄₁₇

Solution

Pressure exerted by hydrogen will be proportional to its mole fraction.

Mole fraction of H₂=\(\frac{\frac{w}{2}}{\frac{w}{16}+\frac{w}{2}}=\frac{8}{9}\)

The volume-temperature graphs of a given mass of an ideal gas at constant pressure are shown below.

What is the correct order of pressures ?

p₁> p₃> p₂
p₁> p₂> p₃
p₂> p₃> p₁
p₂> p₁> p₃

Solution

From the graph we can see the correct order of pressures
p₁> p₃> p₂

Dominance of strong repulsive forces among the molecules of the gas (Z= compressibility factor):

Depends on Z and indicated by Z= 1
Depends on Z and indicated by Z> 1
Depends on Z and indicated by Z< 1
Is independent of Z.

Solution

Repulsive force will decrease the compressibility factor i.e. so, value of Z > 1 as

\(Z=\frac{PV}{RT}\)

Due to repulsion value of PV will be greater than RT so Z> 1.

A gas diffuse 1/5 times as fast as hydrogen. Its molecular weight is

50
25
25\(\sqrt{2}\)
5\(\sqrt{2}\)

Dalton’slaw of partial pressure will not apply to which of the following mixture of gases

H₂ and SO₂
H₂ and Cl₂
H₂ and CO₂
CO₂ and Cl₂

Solution

Because H₂ & Cl₂ gases may react with each other to produce HCl gas hence Dalton’s law is not applicable.

When a substance is dissolved in a solvent, the vapour pressure of solvent decreases. It brings:

adecrease in boiling point of solution
an increase in boiling point ofthe solution
adecrease in freezing point of the solution
an increase in freezing point of the solution

Solution

When vapour pressure decreases, boiling point increases correspondingly.

If three unreactive gases having partial pressures P_A, P_B and P_C and their moles are 1, 2 and 3 respectively then their total pressure will be

P=P_A+P_B+P_C
\(P=\frac{P_{A}+P_{B}+P_{C}}{6}\)
\(P=\frac{\sqrt{P_{A}+P_{B}+P_{C}}}{3}\)
None of these

Pure hydrogen sulphide is stored in a tank of 100 litre capacity at 20° Cand 2atm pressure. The mass of the gas will be

34 g
340 g
282.68 g
28.24 g

Solution

\(n=\frac{PV}{RT}=\frac{m}{M}\)

\(m=\frac{MPV}{RT}=\frac{34\times 2\times 100}{0.082\times 293}=282.68gm\)

If P is pressure and ρ is density of a gas, then P and ρ are related as

P ∝ ρ
P ∝(1/ρ)
P ∝ ρ²
P ∝(1/ρ²)

Solution

\(P\alpha \frac{1}{V}\: and\: \frac{m}{v}=\rho ;\frac{1}{V}=\frac{\rho }{m}\)

So,P∝^ρ⁄_mi.e. Pressure ∝ ρ

16 g of oxygen and 3g of hydrogen are mixed and kept at 760mm of Hg pressure and 0° C. The total volume occupied by the mixture will be nearly

22.4 litres
33.6 litres
448 litres
44800 ml

Solution

n of O₂=¹⁶⁄₃₂=¹⁄₂

n of H₂=³⁄₂

Total no.of moles=³⁄₂+¹⁄₂=2

\(V=\frac{nRT}{P}=\frac{2\times .082\times 273}{1}=44.8lit=44800\: ml\)

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