Gaseous States Test - 13

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Gaseous States Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

Pressure of a mixture of 4g of O₂ and 2g of H₂ confined in a bulb of 1 litre at 0^oC is:

A
25.215 atm

B
31.205 atm

C
45.215 atm

D
15.210 atm

Solution
Question 2
1 / -0

A sample of gas occupies 100 mL at 27^∘C and 740 mm pressure. When its volume is changed to 80 mL at 740 mm pressure, the temperature of the gas will be :

A
21.6^∘C

B
240^∘C

C
−33^∘C

D
89.5^∘C

Solution
Question 3
1 / -0

A sample of a given mass of gas at a constant temperature occupies 95cm3 under a pressure of 9.962×10⁴Nm². At the same temperature its volume at a pressure of 10.13×10⁴Nm⁻² is:

A
190cm³

B
93.42 cm³

C
46.5 cm³

D
47.5 cm³

Solution

According to boyle's law At constant temperature
Question 4
1 / -0

The density of a gas A is three times of a gas B,if the molecular mass of A is M, the molecular mass of B is

A
3M

B
√3 M

C
M/3

D
M/√3

Solution
Question 5
1 / -0

Pure hydrogen sulphide is stored in a tank of 100 litre capacity at 20^∘C and 2 atm pressure. The mass of the gas will be :

[S = 32,H = 1].

A
34 g

B
340 g

C
282.4 g

D
28.24 g

Solution

Here using ideal gas equation,

given :

V = 100L

P = 2atm

M(molar mass) = 34gmol⁻¹

T = 293K

R = 0.0821L atm mol⁻¹T⁻¹

Solution: P×V=n×R×T

Substituting the given values,

n = 8.31mol

But, n = M/W

W = 8.31 × 34

W = 282.54g
Question 6
1 / -0

For an ideal gas, a number of moles per liter in terms of its pressure P, gas constant R and temperature T is:

A
PT/R

B
PRT

C
P/RT

D
RT/P

Solution

According to Ideal Gas Equation, PV = nRT

P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature.

Therefore number of moles per litre,

n/v = P/RT
Question 7
1 / -0

If 2 moles of an ideal gas at 546K occupy volume 44.8L, then pressure must be :

A
2atm

B
3atm

C
4atm

D
1atm

Solution

From gas equation, PV = nRT

Given that :

V = 44.8L
n = 2
R = 0.0821 L atm/mol/K
T = 546K
Question 8
1 / -0

Select one correct statement. In the gas equation, PV = nRT

A
n is the number of molecules of a gas.

B
n moles of the gas have a volume V.

C
V denotes volume of one mole of the gas.

D
P is the pressure if the gas when only one mole of gas is present.

Solution

PV = nRT

where P= pressure of the gas;

V=volume of the gas;

n= Number of Moles;

T=Absolute temperature;

R=Ideal Gas constant also known as Boltzmann Constant = 0.082057 L atm K⁻¹ mol⁻¹.

Here V ∝ n

Hence, V is the volume of n moles of gas
Question 9
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A gas can be liquefied ______.

A
above its critical temperature

B
at its critical temperature

C
below its critical temperature

D
at any temperature

Solution

A gas can be liquefied below its critical temperature.

For example, the critical temperature T_c for CO_2, O₂ and H₂ are 31.1 ⁰C, −118.8 ⁰C and −240 ⁰C respectively.

Thus, CO₂ gas cannot be liquefied above 31.1⁰C however high the pressure may be applied.
Question 10
1 / -0

At 27^∘C, the ratio of rms velocities of ozone to oxygen is

A

B

C

D
0

Solution