Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Equal masses of $$N_{2}$$ and $$O_{2}$$ gases are filled in vessel A and B. The volume of vessel B is double of A. The ratio of pressure in vessel A and B will be

A
16:7

B
16:14

C
32:7

D
32:28

Solution

The mathematical form of the Ideal Gas Law is:
PV = nRT and n = m/MW
Where:P - pressure
V - volume
n - number of moles
T - temperature
m - mass
MW - Molecular Weight
R - ideal gas constant
So we can write,
$$\dfrac{P_a}{
P_b}$$ =$$\dfrac{M_b \times V_b}{ M_a \times V_a}$$
here, $$V_b = 2 V_a $$and$$ M_a$$ = atomic wt of $$O_2$$ = 8 and $$M_b$$ = atomic wt of $$N_2$$ = 7
Putting these values we get,
$$P_a/ P_b$$ = 16/7
Question 2
1 / -0

If the pressure of a gas contained in a closed vessel increases by x% when heated by 1$$^{0}$$C , its initial temperature is

A
$$\frac{100}{x}$$ Kelvin

B
$$\frac{100}{x}$$ Celsius

C
$$\frac{x+100}{x}$$ Kelvin

D
$$\frac{100-x}{x}$$ Celsius
Solution
Let the initial pressure and temperature be P and T.
From Ideal Gas Law,
$$P\propto T$$
$$\implies \dfrac{(1+\dfrac{x}{100})P}{P}=\dfrac{T+1}{T}$$
$$\implies T=\dfrac{100}{x}K$$
Question 3
1 / -0

The pressure of a certain mass of gas at 27$$^{0}$$C is 84cm of Hg. If 25% of the gas is now introduced into the same vessel at the same temperature, the final pressure of the gas will be in cm of Hg

A
105

B
100

C
95

D
90

Solution

Pressure=$$h\rho g=\dfrac{nRT}{V}$$
$$\implies h\propto n\propto m$$
Therefore $$\dfrac{h_2}{h_1}=\dfrac{1.25m_{0}}{m_0}$$
$$\implies h_2=1.25h_1=105cm$$
Question 4
1 / -0

A partially inflated balloon contains 500 m$$^{3}$$ of helium at 27$$^{0}$$ C and 1 atm pressure. The volume of the helium at an altitude of 6000 m, where the pressure is 0.5atm and the temperature is -3$$^{0}$$C is

A
22.4 lit

B
11.2 lit

C
900 m$$^{3}$$

D
50 m$$^{3}$$

Solution

Assuming helium to be an ideal gas, and using relation:
$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$

$$V_2=V_1\dfrac{P_1T_2}{P_2T_1}=900m^3$$
Question 5
1 / -0

28gm of $$N_2$$ gas is contained in a flask at a pressure of 10atm and at a temperature of 57$$^{0}$$C. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to 27$$^{0}$$C. The quantity of N$$_{2}$$ gas that leaked out is

A
$$\frac{5}{63}gm$$

B
$$\frac{63}{5}gm$$

C
$$\frac{11}{20}gm$$

D
$$\frac{20}{11}gm$$

Solution

Since the volume of the flask remains the same, using Ideal Gas Equation we claim,
$$n\propto \dfrac{P}{T}$$
$$\implies \dfrac{n_2}{n_1}=\dfrac{P_2T_1}{P_1T_2}$$
$$\implies n_2=\dfrac{11}{20}n_1$$
Therefore, fraction of gas leaked out is $$(1-\dfrac{11}{20})=\dfrac{9}{20}$$
Thus, amount of gas leaked out=$$\dfrac{9}{20}\times 28gm=\dfrac{63}{5}gm$$
Question 6
1 / -0

In a given species of tobbaco there is 0.1 mg of virus per c.c. The mass of virus is $$4 \times 10^{7}$$ Kg per kilomol. The number of the molecules of virus present in 1 c.c. will be

A
$$10^{13}$$

B
$$10^{11}$$

C
$$1.5 \times 10^{12}$$

D
$$10^{14}$$

Solution

$${ m }_{ v }=\dfrac { M }{ { N }_{ A } } =\dfrac { 4\times { 10 }^{ 7 } }{ 6\times { 10 }^{ 23 }\times { 10 }^{ 3 } } \\ mass\quad of\quad virus\quad per\quad cc=0.1\times { 10 }^{ -6 }kg\\ $$
Number of molecules in cc
$$ n=\dfrac { { m }_{ v } }{ { N }_{ A } } =\dfrac { 0.1\times { 10 }^{ -6 }\times 6\times { 10 }^{ 23 }\times { 10 }^{ 3 } }{ 4\times { 10 }^{ 7 } } \\ n=1.5\times { 10 }^{ 12 }$$
Question 7
1 / -0

$$Assertion$$: gases are characterised with two coefficients of expansion
$$Reason$$: when heated both volume and pressure increase with the rise in temperature

A
Both assertion (A) and reason (R) are correct and R gives the correct explanation

B
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation

C
A is true but R is false

D
A is false but R is true

Solution

The ideal gas equation $$PV=\eta RT$$ tells us that the pressure and volume are directly proportional to the temperature. The coefficients of expansion for pressure is $$\gamma_p$$ and that of volume is $$\alpha_v$$.
Question 8
1 / -0

Assertion: Gasses obey Boyle's law at high temperature and low pressure only.
Reason: At low pressure and high temperature, gasses would behave like ideal gases.

A
Both assertion (A) and reason (R) are correct and R gives the correct explanation

B
Both assertion (A) and reason (R) are correct but R does give the correct explanation

C
A is true but R is false

D
A is false but R is true

Solution

Due to non idealities gases obeys vander waals equation instead of ideal equation and at high temperature and low pressure vander waals equation becomes ideal gas equation by taking approximations. therefore, at high temperatur and low pressure gases obeys boyles law and behaves ideally.
assertion is true reason is true and reason is correct explanation of assertion.
option(A)
Question 9
1 / -0

Follwing operation are carried out on a sample of ideal gas initially at pressure P volume V and kelvin temperature T.
a) At constant volume, the pressure is increased fourfold.
b) At constant pressure, the volume is doubled
c) The volume is doubled and pressure halved.
d) If heated in a vessel open to atmosphere, one-fourth of the gas escapes from the vessel.
Arrange the above operations in the increasing order of final temperature.

A
a,b,c,d

B
c,b,a,d

C
b,a,d,c

D
c,d,b,a

Solution

ideal gas equation
$$PV = nRT$$ eq(1)
where
$$P$$ = Pressure
$$V$$ = Volume
$$n$$ = number of moles
$$R$$ = universal gas constant
$$T$$ = Temperature

suppose initial values are $${P}_{1},{V}_{1},{n}_{1},{T}_{1}$$ and
final values are $${P}_{2},{V}_{2},{n}_{2},{T}_{2}$$

so
$${P}_{1}{V}_{1}={n}_{1}R{T}_{1}$$ eq(2)
$${P}_{2}{V}_{2}={n}_{2}R{T}_{2}$$ eq(3)

(a) At constant volume, the pressure is increased fourfold
$${V}_{2} = {V}_{1}$$
$${P}_{2} = 4 {P}_{1}$$
$${n}_{2} = {n}_{1}$$
$$\dfrac{eq(3)}{eq(2)}$$
$$\dfrac{{T}_{2}}{{T}_{1}} = 4$$
$${T}_{2} = 4 {T}_{1}$$

(b) At constant pressure, the volume is doubled
$${V}_{2} = 2 {V}_{1}$$
$${P}_{2} = {P}_{1}$$
$${n}_{2} = {n}_{1}$$
$$\dfrac{eq(3)}{eq(2)}$$
$$\dfrac{{T}_{2}}{{T}_{1}} = 2$$
$${T}_{2} = 2 {T}_{1}$$

(c) The volume is doubled and pressure halved.
$${V}_{2} = 2 {V}_{1}$$
$${P}_{2} =\dfrac{1}{2} {P}_{1}$$
$${n}_{2} = {n}_{1}$$
$$\dfrac{eq(3)}{eq(2)}$$
$$\dfrac{{T}_{2}}{{T}_{1}} = 1$$
$${T}_{2} = {T}_{1}$$

(d) If heated in a vessel open to atmosphere, onefourth of the gas escapes from the vessel.

for open vessel pressure and volume will be constant.

$${V}_{2} = {V}_{1}$$
$${P}_{2} = {P}_{1}$$

one-fourth of gas escaped from vessel, so three-fourth gas left
$${n}_{2} = \dfrac{3}{4}{n}_{1}$$

$$\dfrac{\dfrac{3}{4} {T}_{2}}{{T}_{1}} = 1$$

$${T}_{2} = \dfrac{4}{3}{T}_{1}$$

arranging final temperature in increasing order

(c),(d),(b),(a)
Question 10
1 / -0

Assertion:With increase in temperature, the pressure of given gas increases
Reason:Increase in temperature causes decrease in no. of collision of molecules with walls of container.

A
Both assertion (A) and reason (R) are correct and R gives the correct explanation

B
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation

C
A is true but R is false

D
A is false but R is true

Solution

With increase in temperature, the pressure of given gas increases. This is because the increase in temperature causes increase in no. of collision of molecules with the walls of the container.

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Re-Attempt Weekly Quiz Competition

Thermal Properties of Matter Test - 30

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Thermal Properties of Matter Test - 30

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SHARING IS CARING

Question 1

16:7

16:14

32:7

32:28

Solution

Question 2

$$\frac{100}{x}$$ Kelvin

$$\frac{100}{x}$$ Celsius

$$\frac{x+100}{x}$$ Kelvin

$$\frac{100-x}{x}$$ Celsius

Solution

Question 3

105

100

95

90

Solution

Question 4

22.4 lit

11.2 lit

900 m$$^{3}$$

50 m$$^{3}$$

Solution

Question 5

$$\frac{5}{63}gm$$

$$\frac{63}{5}gm$$

$$\frac{11}{20}gm$$

$$\frac{20}{11}gm$$

Solution

Question 6

$$10^{13}$$

$$10^{11}$$

$$1.5 \times 10^{12}$$

$$10^{14}$$

Solution

Question 7

Both assertion (A) and reason (R) are correct and R gives the correct explanation

Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation

A is true but R is false

A is false but R is true

Solution

Question 8

Both assertion (A) and reason (R) are correct and R gives the correct explanation

Both assertion (A) and reason (R) are correct but R does give the correct explanation

A is true but R is false

A is false but R is true

Solution

Question 9

a,b,c,d

c,b,a,d

b,a,d,c

c,d,b,a

Solution

Question 10

Both assertion (A) and reason (R) are correct and R gives the correct explanation

Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation

A is true but R is false

A is false but R is true

Solution

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