States of Matter : Gases & Liquids Test

Result

States of Matter : Gases & Liquids Test - 6

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

A sample of an ideal gas occupies 20 L under a pressure of 0.5 atm and under isothermal condition. On increasing its pressure to 1.0 atm, its volume will be

A
20L

B
10L

C
5 L

D
2.5L

Solution

Since the system is isothermal, so temperature is constant.
According to Boyle's lawP₁V₁= P₂V₂
20×0.5=1×V₂
10=V₂
Question 2
4 / -1

A 10.0 cm column of air is trapped by a column of Hg 4.00 cm long in a capillary tube of uniform bore when the tube is held horizontally at 1 atm. Length of the air column when the tube is held vertically with the open end up is

A
9.50 cm

B
10.52 cm

C
3.53cm

D
4.61 cm

Solution

V1 = 10xcm³
V2 = ?
P1 = 1atm = 760mmHg
P2 = (760 + 4)mmHg
P1V1 = P2V2
V2 = P1V1 / P2
= 760 × 10x / 764
= 9.5x
Therefore,
= 9.5x / x = 9.5
Question 3
4 / -1

Volume of an ideal gas is to be decreased by 10% by increase of pressure by x% under isothermal condition. Thus, x is

A
100/9

B
9/100

C
10

D
1/10

Solution

Applying Boyle 's law P1V1=P2V2 as temp is constant.
So let initial pressure =P initial volume =V
Now final volume V2=V-10%of V =9V/10
PV=P2×9V/10 so P2=10P/9
Therefore increment in Pressure=P/9P×100%=100/9
Question 4
4 / -1

Density of an ideal gas at 298 K and 1.0 atm is found to be 1.25 kg m^-3. Density of the gas at 1.5 atm and at 298 K is

A
1.25 kg m^-3

B
1.875 kg m^-3

C
1.00 kg m^-3

D
1.20 kg m^-3

Solution

We know ideal gas equation,
PV=nRT.
n= m(mass)/M(molecular weight).
PV=m/M RT.
PM=m/v RT.
density (d)=m(mass)/v (volume).
PM=dRT.
d=PM/RT.
here, density is directly proportional to pressure and inversely proportional to temperature...
d₁/d₂=P₁ T₂ / P₂ T₁.
1.25/d₂= 1×298/1.5×298.
1.25/d₂=1/1.5.
d₂= 1.25×1.5.
d₂=1.875
Question 5
4 / -1

An ideal gas is at 300 K and 1 atm. Then volume remains constant if

A
temperature is doubled and pressure is halved

B
temperature is halved and pressure is doubled

C
temperature is doubled and pressure is doubled

D
only temperature is doubled

Solution

We know that
pV = nRT
Or V = nRT/p
If we increase both T and P 2 times, only then the volume remains constant.
Question 6
4 / -1

The pressure of a 1:4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen?

A
0.8 × 10⁵ atm

B
0.008 Nm^–2

C
8 × 10⁴ Nm^–2

D
0.25 atm

Solution

Let the number of moles of dihydrogen and dioxygen be 1 and 4.
Question 7
4 / -1

At 298 K, which of the following gases has the lowest average molecular speed?

A
CO₂ at 0.20 atm

B
He at 0.40 atm

C
CH₄ at 0.80 atm

D
NO at 1.00 atm

Solution

Under the same conditions the average kinetic energy of the gases should be equal.
speed of gas1/speed of gas 2 = square root of (molar mass of gas2/molar mass of gas 1)
So gases with small molar mass will move quicker
eg speed of N₂/speed of CO₂ = sq. rt (44/28) = 1.25
speed of N₂/speed of F₂ = sq rt. (38/28) = 1.16
So order is N₂, F₂ and CO₂
then we concluded right ans. is (A)
Question 8
4 / -1

Charle's law defines value of 'k' as

A
V/n

B
V/V

C
V/P

D
V/T

Solution

According to Charles law,at constant pressure, V∝T
Or V/T = k
Question 9
4 / -1

By what ratio will the average velocity of the molecules in a gas change when the temperature is raised from 50°C to 200°C?

A
1.21/1

B
1.46/1

C
2/1

D
4/1

Solution

This is the correct answer. If we take v₂₀₀/v₅₀, then we will get option b. But that is not according to the question.
Question 10
4 / -1

For Charles’ law, graphical representation between log V and log T under isobaric condition is shown below for equation V = kT.

Q. Relation between p₁, p₂and p₃ is

A
p₁ = p₂ = p₃

B
P₁ < P₂ < P₃

C
p₃ < p₂ < P₁

D
P₁ < P₃ < P₂

Solution

From ideal gas equation, we have
pV = nRT
Or V = nRT/p
Taing log on both sides
log V = log nRT - log p
Or y = mx - c
Let us make a point constant on time axis like this

Now we have the temperature constant.So to have maximum value of log V, we need to have minimum value of log p. From the graph we can see that we have least pressure for p3 . So the order will be p₃<p₂<p₁.
Question 11
4 / -1

For Charles’ law, graphical representation between log V and log T under isobaric condition is shown below for equation V = kT

Q. Volume V = constant k, when

A
T = 0^oc

B
T = 273^oc

C
T = 0K

D
T = -272^oc

Solution

-272deg.
For C = 1 K
V / 1 = K
V = K
Question 12
4 / -1

Graphicaily Boyle’s law can be represented by various options depending on coordinates. Match the graphs in Column t with their coordinates in Column II and select the correct alternate from the codes given below.

A
a

B
b

C
c

D
d

Solution

From Boyle’s law;
p ∝ 1/V ⇒ Graph i (Straight line passing through origin)
Or pV = Constant ⇒ Graph ii (rectangular hyperbola)
Taking log on both side;
log(pV) = logK
logp + logV = k’
logp = -logV = k’ (y = -mx+c) ⇒ Graph iii
log p = log V-1 + k’
Logp = log(1/V) + k’ ⇒ Graph iv
So option c is correct
Question 13
4 / -1

Direction (Q. Nos. 13 and 14) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive).

Isothermally compressibility of an ideal gas is

Q. What is the value of a at 0.2 bar?

A
5

B
6

C
7

D
9

Solution

α = -1/V(dV/dP)_T
pV = RT;
V = RT/p
dV/dp = -RT/p²
α = -1/(RT/p)×(-RT/p²)
α = 1/p
Or α = (since p = 0.2)
Question 14
4 / -1

Cubic expansion coefficient (β) for an ideal gas is

Q. Its value at 250 K is x * 10^-3 K ^-1. What is value of x?

A
8

B
7

C
5

D
4

Solution