Heat and Thermodynamics Questions — NEET Physics

Q1

A flask contains argon and chlorine in the ratio of

2: 1

by mass. The temperature of the mixture is

27^{\circ} \mathrm{C}

. The ratio of root mean square speed of the molecules of the two gases

\left(\dfrac{V_{\mathrm{rms}}^{\mathrm{Ar}}}{V_{\mathrm{rms}}^{\mathrm{Cl}}}\right)

is: (Atomic mass of argon

=40.0 \mathrm{u}

and molecular mass of chlorine

=70.0 \mathrm{u}

)

A

\dfrac{\sqrt{7}}{2}

B

\dfrac{7}{4}

C

\dfrac{7}{2}

D

\dfrac{2}{\sqrt{7}}

Correct Answer

Option A

Solution

$v_{\mathrm{rms}}=\sqrt{\dfrac{3 R T}{M}}$ For same temperature, $v_{\text{rms }} \propto \dfrac{1}{\sqrt{M}}$

\frac{V_{\mathrm{rms}}^{\mathrm{Ar}}}{V_{\mathrm{rms}}^{\mathrm{Cl}}}=\sqrt{\frac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}}=\sqrt{\frac{70}{40}}=\frac{\sqrt{7}}{2}

Q2

An electric heater supplies heat to a system at a rate of 100 W . If the system performs work at a rate of

75 \mathrm{~J} / \mathrm{s}

, then the rate at which internal energy increases will be:

A 75 W

B 100 W

C 125 W

D 25 W

Correct Answer

Option D

Solution

Using $1^{\text{st }}$ law of thermodynamics for electric heater,

\begin{aligned} & Q=\Delta U+W \\ & \Rightarrow \quad \frac{d Q}{d t}=\frac{d(\Delta U)}{d t}+\frac{d W}{d t} \\ & \Rightarrow \quad 100 \mathrm{~W}=\frac{d(\Delta U)}{d t}+75 \mathrm{~W} \\ & \therefore \quad \frac{d(\Delta U)}{d t}=25 \mathrm{~W} \end{aligned}

Q3

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity

2 K

while that in the middle has thermal conductivity

K

. The left end of the combination is maintained at temperature

3 T

and the right end at

T

. The rods are thermally insulated from outside. In steady state, temperature at the left junction is

T_1

and that at the right junction is

T_2

. The ratio

T_1 / T_2

is

A

\dfrac{5}{3}

B

\dfrac{5}{4}

C

\dfrac{3}{2}

D

\dfrac{4}{3}

Correct Answer

Option A

Solution

First, consider the equivalent thermal resistance $R_{\text{eq}}$ of the series arrangement: $R_{\text{eq}} = R_1 + R_2 + R_3$ Calculating each resistance: $R_1 = \dfrac{1}{2KA}, \quad R_2 = \dfrac{1}{KA}, \quad R_3 = \dfrac{1}{2KA}$ Summing these gives: $R_{\text{eq}} = \dfrac{1}{2KA} + \dfrac{1}{KA} + \dfrac{1}{2KA} = \dfrac{2}{KA}$ In a series arrangement, the rate of heat flow is constant.

Thus: $\dfrac{3T - T_1}{R_1} = \dfrac{3T - T}{R_{\text{eq}}}$ Substituting the resistances: $\dfrac{(3T - T_1) \cdot 2KA}{I} = \dfrac{2T \cdot KA}{2I}$ Solving the equation yields: $6T - 2T_1 = T \implies 2T_1 = 5T \implies T_1 = \dfrac{5T}{2} \quad \ldots (1)$ Next, examine the heat flow rate in the third section: $\dfrac{T_2 - T}{R_3} = \dfrac{3T - T}{R_{\text{eq}}}$ Substitute: $\dfrac{(T_2 - T) \cdot 2KA}{I} = \dfrac{2T \cdot KA}{2I}$ Solving gives: $2T_2 - 2T = T \implies 2T_2 = 2T + T \implies T_2 = \dfrac{3T}{2} \quad \ldots (2)$ By substituting equations (1) and (2) into the ratio: $\dfrac{T_1}{T_2} = \dfrac{5T/2}{3T/2} = \dfrac{5}{3}$ Thus, the ratio $\dfrac{T_1}{T_2}$ is $\dfrac{5}{3}$ .

Q4

A container has two chambers of volumes

V_1=2

litres and

V_2=3

litres separated by a partition made of a thermal insulator. The chambers contain

n_1=5

and

n_2=4

moles of ideal gas at pressures

p_1=1 \mathrm{~atm}

and

p_2=2 \mathrm{~atm}

, respectively. When the partition is removed, the mixture attains an equilibrium pressure of

A 1.4 atm

B 1.8 atm

C 1.3 atm

D 1.6 atm

Correct Answer

Option D

Solution

To find the equilibrium pressure of the system after the partition is removed, we can apply the principle of conservation of moles (or the ideal gas law in combined volumes).

Initially, we have: Chamber 1: $P_1 = 1 \, \text{atm}, \, V_1 = 2 \, \text{L}$ Chamber 2: $P_2 = 2 \, \text{atm}, \, V_2 = 3 \, \text{L}$ The total pressure after the partition is removed and the gases mix can be calculated using: $P_{\text{final}} = \dfrac{P_1 V_1 + P_2 V_2}{V_1 + V_2}$ Substituting the given values: $P_{\text{final}} = \dfrac{1 \times 2 + 2 \times 3}{2 + 3} = \dfrac{2 + 6}{5} = \dfrac{8}{5}$ This gives us: $P_{\text{final}} = 1.6 \, \text{atm}$ Therefore, the equilibrium pressure when the gases mix is $1.6 \, \text{atm}$ .

Q5

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature

27^{\circ} \mathrm{C}

. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given,

R=\dfrac{100}{12} \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

, and molecular mass of

\mathrm{O}_2=32,1

atm pressure

=1.01 \times 10^5 \mathrm{~N} / \mathrm{m}

]

A 0.116 kg

B 0.156 kg

C 0.125 kg

D 0.144 kg

Correct Answer

Option A

Solution

To find the mass of oxygen withdrawn from the cylinder, we start by calculating the number of moles left in the cylinder after some oxygen is withdrawn.

We use the ideal gas law in the form: $n = \dfrac{PV}{RT}$ Substituting the given values: $P = 11$ atm converts to $11 \times 1.01 \times 10^5 \, \text{N/m}^2$ $V = 30 \, \text{liters} = 30 \times 10^{-3} \, \text{m}^3$ $R = \dfrac{100}{12} \, \text{J/mol K}$ $T = 27^\circ \text{C} = 300 \, \text{K}$ We calculate the moles after oxygen has been withdrawn: $n = \dfrac{12 \times 1.01 \times 10^5 \, \text{N/m}^2 \times 30 \times 10^{-3} \, \text{m}^3}{\left(\dfrac{100}{12}\right) \times 300}$ Simplifying the expression: $n = \dfrac{12 \times 1.01 \times 12}{10} = 14.54 \, \text{moles}$ Next, determine the moles of oxygen removed: $\text{Moles removed} = 18.20 - 14.54 = 3.656 \, \text{moles}$ Finally, convert the moles removed into mass: $\text{Mass removed} = 3.656 \times 32 = 116.99 \, \text{g} = 0.116 \, \text{kg}$ Thus, the mass of the oxygen withdrawn from the cylinder is approximately 0.116 kg.

Q6

Two gases

A

and

B

are filled at the same pressure in separate cylinders with movable pistons of radius

r_A

and

r_B

, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas

A

and

B

are displaced by 16 cm and 9 cm , respectively. If the change in their internal energy is the same, then the ratio

\dfrac{r_A}{r_B}

is equal to

A

\dfrac{2}{\sqrt{3}}

B

\dfrac{\sqrt{3}}{2}

C

\dfrac{4}{3}

D

\dfrac{3}{4}

Correct Answer

Option D

Solution

The problem involves two gases, $A$ and $B$ , at the same pressure in separate cylinders with movable pistons of radii $r_A$ and $r_B$ .

Equal amounts of heat are supplied to both cylinders reversibly under constant pressure, displacing the pistons by 16 cm for gas $A$ and 9 cm for gas $B$ .

Given that the change in their internal energy is the same, we need to find the ratio $\dfrac{r_A}{r_B}$ .

First, we apply the first law of thermodynamics: $\Delta Q = \Delta U + P \Delta V$ Since $\Delta Q$ and $\Delta U$ are the same for both gases, their work done $W_A$ and $W_B$ is also equal.

Therefore: $(P \Delta V)_A = (P \Delta V)_B$ With constant pressure $P$ , the relationship becomes: $A_A d_A = A_B d_B$ where $A$ is the area of the piston.

We know: $\pi r_A^2 d_A = \pi r_B^2 d_B$ Simplifying, the ratio of the radii is: $\dfrac{r_A}{r_B} = \left(\dfrac{d_B}{d_A}\right)^{\dfrac{1}{2}} = \left(\dfrac{9}{16}\right)^{\dfrac{1}{2}}$ Calculating the square root gives: $\dfrac{r_A}{r_B} = \dfrac{3}{4}$

Q7

According to the law of equipartition of energy, the number of vibrational modes of a polyatomic gas of constant

\gamma=\dfrac{C_p}{C_v}

is (

C_P

where

C_V

are the specific heat capacities of the gas at constant pressure and constant volume, respectively):

A

\dfrac{4+3 \gamma}{\gamma-1}

B

\dfrac{3+4 \gamma}{\gamma-1}

C

\dfrac{4-3 \gamma}{\gamma-1}

D

\dfrac{3-4 \gamma}{\gamma-1}

Correct Answer

Option C

Solution

A polygamic gas has 3 translational, 3 rotational and

f

vibration modes

\begin{aligned} & U=\frac{3}{2} k_B T+\frac{3}{2} k_B T+f k_B T \\ & U=(3+f) k_B T \\ & C_V=(3+f) R \\ & C_p=(4+f) R \\ & \frac{C_p}{C_v}=\frac{4+f}{3+f}=\gamma \\ & 4+f=3 \gamma+f \gamma \\ & 4-3 \gamma=f(\gamma-1) \Rightarrow f=\frac{4-3 \gamma}{\gamma-1} \end{aligned}

Q8

A thermodynamic system is taken through the cycle

abcda

. The work done by the gas along the path

b c

is:

A Zero

B 30 J

C

-

90 J

D

-

60 J

Correct Answer

Option A

Solution

Path

b c

is an isochoric process.

\therefore \quad

Work done by gas along path

b c

is zero.

Q9

The following graph represents the

T

-

V

curves of an ideal gas (where

T

is the temperature and

V

the volume) at three pressures

P_1, P_2

and

P_3

compared with those of Charles's law represented as dotted lines. Then the correct relation is :

A

P_3>P_2>P_1

B

P_1>P_3>P_2

C

P_2>P_1>P_3

D

P_1>P_2>P_3

Correct Answer

Option D

Solution

At same temperature, curve with higher volume corresponds to lower pressure.

\begin{aligned} & V_3>V_2>V_1 \\ & \Rightarrow P_1>P_2>P_3 \end{aligned}

(We draw a straight line parallel to volume axis to get this)

Q10

For the given cycle, the work done during isobaric process is:

A 200 J

B Zero

C 400 J

D 600 J

Correct Answer

Option D

Solution

\mathrm{AB}

is isobaric process

\begin{aligned} & \mathrm{W}_{\mathrm{AB}}=\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & \mathrm{W}_{\mathrm{AB}}=3 \times 10^2(3-1) \\\\ & \mathrm{W}_{\mathrm{AB}}=3 \times 100 \times 2 \\\\ & \mathrm{~W}_{\mathrm{AB}}=600 \mathrm{~J} \end{aligned}