Free Objective Test 01 Practice Test - 11th and 12th

Question 1

Two thermally insulated vessels $1$ and $2$ are filled with air at temperature ( $T_{1}$ , $T_{2}$ ), volume ( $V_{1}$ , $V_{2}$ ) and pressure ( $P_{1}$ , $P_{2}$ ) respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

$T_{1} + T_{2}$

$\frac{T_{1} + T_{2}}{2}$

$\frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1})}$

$\frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{P_{1} V_{1} T_{1} + P_{2} V_{2} T_{2})}$

SOLUTION

Solution : C

According to the kinetic theory, the average kinetic energy(KE) per molecule of a gas

= $\frac{3}{2} K T$ . Let $n_{1}$ and $n_{2}$ be the number of moles of air in vessels 1 and 2 respectively.

Before mixing, the total KE of molecules in the two vessels is

$E_{1} = \frac{3}{2} n_{1} k T_{1} + \frac{3}{2} n_{2} k T_{2}$

= $\frac{3}{2} k (n_{1} T_{1} + n_{2} T_{2})$

After mixing, the total KE of molecules is

$E_{2} = \frac{3}{2} (n_{1} + n_{2}) k T$

Where T is the temperature when equilibrium is established. Since there is no loss of energy

(Because the vessels are insulated), $E_{2} = E_{1}$ or

$\frac{3}{2} (n_{1} + n_{2}) k T = \frac{3}{2} k (n_{1} T_{1} + n_{2} T_{2})$

or $T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

Now $P_{1} V_{1} = n_{1} R T_{1}$ and $P_{2} V_{2} = n_{2} R T_{2}$ which give

$n_{1} = \frac{P_{1} V_{1}}{R T_{1}}$ and $n_{2} = \frac{P_{2} V_{2}}{R T_{2}}$

Using these in Eq. (1) and simplifying, we get

$T = \frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{(P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1})}$

Question 2

A vessel contains $1$ mole of $O_{2}$ gas (molar mass $32$ ) at a temperature $T$ . The pressure of the gas is $P$ . An identical vessel containing one mole of $H e$ gas (molar mass $4$ ) at a temperature $2 T$ has a pressure of

$\frac{P}{8}$

SOLUTION

Solution : C

For a gas, $P V = n R T$ . Hence

$(P)_{O_{2}} = \frac{(1 m o l e) R T}{V}$

and $(P)_{H e} = \frac{(1 m o l e) R (2 T)}{V}$

$∴ \frac{(P)_{H e}}{(P)_{O_{2}}} = 2$ or $(P)_{H e} = 2 (P)_{O_{2}}$

Question 3

An enclosure of volume $V$ contains a mixture of $8 g$ of oxygen, $14 g$ of nitrogen and $22 g$ of carbon- dioxide at absolute temperature $T$ . The pressure of the mixture of gases is ( $R$ is universal gas constant)

$\frac{R T}{V}$

$\frac{3 R T}{2 V}$

$\frac{5 R T}{4 V}$

$\frac{7 R T}{5 V}$

SOLUTION

Solution : C

The pressure exerted by a gas is given by

$P = \frac{n R T}{V}$

= $\frac{m a s s}{m o l e c u l a r w e i g h t} \times \frac{R T}{V}$

Pressure exerted by oxygen $P_{1} = \frac{8}{32} \frac{R T}{V} = \frac{1}{4} \frac{R T}{V}$

Pressure exerted by oxygen $P_{2} = \frac{14}{28} \frac{R T}{V} = \frac{1}{2} \frac{R T}{V}$

Pressure exerted by carbon dioxide

$P_{3} = \frac{22}{44} \frac{R T}{V} = \frac{1}{2} \frac{R T}{V}$

From Dalton's law of partial pressures, the total pressure exerted by the mixture is given by

$P = P_{1} + P_{2} + P_{3}$

= $\frac{1}{4} \frac{R T}{V} + \frac{1}{2} \frac{R T}{V} + \frac{1}{2} \frac{R T}{V}$

= $\frac{5 R T}{4 V}$ ,

which is choice (c)

Question 4

The relation between volume $V$ pressure $P$ and absolute temperature $T$ of an ideal gas is $P V = x T$ where x is a constant. The value of x depends upon

The mass of the gas molecule

The average kinetic energy of the gas molecules

P, V and T

The number of gas molecules in volume V

SOLUTION

Solution : A and D

x= nR where n is the number of moles of the gas

Question 5

A vessel of volume $V$ contains an ideal gas at absolute temperature $T$ and pressure $P$ . The gas is allowed to leak till its pressure falls to $P$ . Assuming that the temperature remains constant during leakage, the number of moles of the gas that have leaked is

$\frac{V}{R T} (P + P^{'})$

$\frac{V}{2 R T} (P + P^{'})$

$\frac{V}{R T} (P - P^{'})$

$\frac{V}{2 R T} (P - P^{'})$

SOLUTION

Solution : C

Number of moles present initially is $n = \frac{P V}{R T}$ . Let $n^{'}$ be the number of moles of the gas that leaked till

the pressure falls to $P^{'}$ . Since volume V of the vessel cannot change and temperature T remains

constant during leakage, we have

$n^{'} = \frac{P^{'} V}{R T}$

$∴$ Number of moles that leaked is

$Δ n = n - n^{'} = \frac{P V}{R T} - \frac{P^{'} V}{R T} = \frac{V}{R T} (P - P^{'})$

So the correct choice is (C)

Question 6

An air bubble doubles its radius on raising from the bottom of water reservoir to be the surface of water in it. If the atmospheric pressure is equal to 10 m of water, the height of water in the reservoir is

A. 10 m

B. 20 m

C. 70 m

D. 80 m

SOLUTION

Solution : C

Accoding to Boyle's law $(P_{1} V_{1})_{b o t t o m}$ = $(P_{2} V_{2})_{t o p}$
(10 + h) $\times$ $\frac{4}{3} π {r_{1}}^{3}$ = 10 $\times$ $\frac{4}{3} π {r_{2}}^{3}$ but $r_{2} = 2 r_{1}$
$∴$ (10 + h) ${r_{1}}^{3}$ = 10 $\times$ 8 ${r_{1}}^{3}$ $∴$ 10 + h = 80 $∴$ h = 70 m

Question 7

If the r.m.s. velocity of a gas at a given temperature (Kelvin scale) is 300 m/s, whathat will be the r.m.s. velocity of a gas having twice the molecular weight and half the temperature on Kelvin scale =

A. 300 m/sec

B. 600 m/sec

C. 75 m/sec

D. 150 m/sec

SOLUTION

Solution : D

$v_{m a x}$ = $\sqrt{\frac{3 R T}{M}} \Rightarrow V_{m a x} \infty \sqrt{\frac{T}{M}}$
$\frac{v_{2}}{v_{1}}$ = $\sqrt{\frac{M_{1}}{M_{2}} \times \frac{T_{2}}{T_{1}}} \sqrt{\frac{1}{2} \times \frac{1}{2}}$ $\Rightarrow v_{2} = \frac{v_{1}}{2} = \frac{300}{2} = 150 m / s e c$

Question 8

The ratio of two specific heats $\frac{C_{p}}{C_{v}}$ of CO is

A. 1.33

B. 1.40

C. 1.29

D. 1.66

SOLUTION

Solution : B

CO is diatomic gas, for diatomic gas
$C_{p} = \frac{7}{2} R$ and $C_{v} = \frac{5}{2} R \Rightarrow γ = \frac{C_{p}}{C_{v}} = \frac{\frac{7 R}{2}}{\frac{5 R}{2}}$ = 1.4

Question 9

The energy of a gas per litre is 300 joules, then its pressure will be

A. 3

\times 10^{5} N m^{2}

B. 6

\times 10^{5} N m^{2}

10^{5} N m^{2}

D. 2

\times 10^{5} N m^{2}

SOLUTION

Solution : D

Energy = 300 J/litre = 300 $\times 10^{3} J / m^{3}$
P= $\frac{2}{3} E = \frac{2 \times 300 \times 10^{3}}{3}$ = 2 $\times 10^{5} N / m^{2}$

Question 10

Pressure versus temperature graphs of an ideal gas are as shown in figure. Choose the wrong statement

A. Density of gas is increasing in graph (i)

B. Density of gas is decreasing in graph (ii)

C. Density of gas is constant in graph (iii)

D. None of these

SOLUTION

Solution : C

(v) $μ$ = $\frac{P M}{R T}$
Density $μ$ remains constant when $\frac{P}{T}$ or volume remains constant.
In graph (i) Pressure is increasing at constant temperature hence volume is decreasing so density is increasing. Graphs (ii) and (iii) volume is increasing hence, density is decreasing. Not that
volume whould had been constant in case the stright line in graph (iii) had passed through origin

Question 11

If pressure of $C O_{2}$ (real gas) in a container is given by P = $\frac{R T}{2 V - b} - \frac{a}{4 V^{2}}$ then mass of the gas in container is

A. 11 gm

B. 22 gm

C. 33 gm

D. 44 gm

SOLUTION

Solution : B

Vander wal's gas equation for $μ$ mole of real gas
(P + $\frac{μ^{2} α}{V^{2}}$ )(V - $μ$ b) = $μ$ RT $\Rightarrow$ P = $\frac{μ R T}{V - μ b} - \frac{μ^{2}}{V^{2}}$
on comparing the given equation with this standard equation we get $μ = \frac{1}{2}$ . Hence $μ$ = $\frac{m}{M} \Rightarrow$ mass of gas
m = $μ M$ = $\frac{1}{2} \times 44$ = 22gm

Question 12

A cylinder of fixed capacity 44.8 litre. contains a monatomic gas at standard temperature and pressure. The amount of heat required to increse the temperature of cylinder by $10^{\circ}$ C will be. (R = universal gas constant)

A. R

B. 10R

C. 20R

D. 30R

SOLUTION

Solution : D

As we know 1 mol of any ideal gas at STP occupies a volume of 22.4 liters.
Hence number of moles of gas $μ = \frac{44.8}{22.4}$ = 2
Since the volume of cylinder is fixed,
Hence $(△ Q)_{V}$ = $μ (C_{V}) △ T = 2 \times \frac{3}{2} R \times 10$ = 30R( $∵ (C_{V})_{m o n o} = \frac{3 R}{2}$ ))

Question 13

A pressure cooker contains air at 1 atm and $30^{\circ}$ C. If the safety value of the cooler blows when the inside pressure $\geq$ 3 atm, then the maximum temperature of the air, inside the cooker can be

90^{\circ}

636^{\circ}

909^{\circ}

363^{\circ}

SOLUTION

Solution : B

Since volume is constant,
Hence $\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}} \Rightarrow \frac{1}{3} = \frac{(273 + 30)}{T_{2}}$
$\Rightarrow$ $T_{2}$ = 909 K = $636^{\circ}$ C

Question 14

One mole of an ideal monatomic gas requires 210 J heat to raise the temperature by 10 K, when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by 10 K then heat required is

A. 238 J

B. 126 J

C. 210 J

D. 350 J

SOLUTION

Solution : B

$(△ Q)_{P}$ = $μ C_{P} △ T$ and $(△ Q)_{V}$ = $μ C_{V} △ T$
$\Rightarrow$ $\frac{(△ Q)_{V}}{(△ Q)_{P}}$ = $\frac{C_{v}}{C_{p}}$ = $\frac{\frac{3}{2} R}{\frac{5}{2} R}$ = $\frac{3}{5}$
[ $∵$ $(C_{V})_{m o n o} = \frac{3}{2} R, (C_{P})_{m o n o} = \frac{5}{2} R$ ]
$\Rightarrow$ $(△ Q)_{V}$ = $\frac{3}{5} \times (△ Q)_{P}$ = $\frac{3}{5} \times 210 = 126 J$

Question 15

From the following V-T diagram we can conclude

P_{1} = P_{2}

P_{1} > P_{2}

P_{1} < P_{2}

D. None of these

SOLUTION

Solution : C

In case of given graph, V and T are related as V = aT + b, where a and b are constants.
From ideal gas equation, PV = $μ R T$
We find P = $\frac{μ R T}{a T + b}$ = $\frac{μ}{a + \frac{b}{T}}$
Since $T_{2} > T_{1}$ , therefore $P_{2} > P_{1}$ .

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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