Figure shows plot of P V / T versus P for 1 . 00 × 10 - 3 kg of oxygen gas at two different temperatures. What is the value of P V / T where the curves meet on the y -axis?

According to the ideal gas equation P V = n R T ⇒ P V T = n R , This indicates the value of P V T is constant for the ideal gas, in the graph the curve for the temperature T 1 is flatter than the other curve because ideal gas behaviour holds good at higher temperature. For the values of P V T , From the equation, P V T = m M R ⇒ P V T = 10 - 3 32 × 10 - 3 × 8 . 314 ⇒ P V T = 0 . 26 J K - 1

An ideal gas having initial pressure P , volume V and temperature T is allowed to expand adiabatically until is volume becomes 5 . 66 V while its temperature falls to T 2 . (i) How many degrees of freedom do gas molecules have? (ii) Obtain the work done by the gas expansion as a function of the initial pressure P and volume V . [Take Take 5 . 66 0 . 4 = 2

For polytropic process T 1 V 1 n - 1 = T 2 V 2 n - 1 ⇒ T V n - 1 = T 2 5 . 66 V n - 1 ⇒ 2 = 5 . 66 n - 1 Taking log both side ℓn 2 = n - 1 ℓn 5 . 66 ⇒ n = 1 . 4 = 1 + 2 f ⇒ f = 5 (i) ∴ Degrees of freedom = 5 (ii) Work done by gas = - P 1 V 1 - P 2 V 2 γ - 1 = P V - P 2 5 . 66 V 1 . 4 - 1 = 12 . 3 PV where P 2 V 2 γ = P 1 V 1 γ ⇒ P 2 = P 1 5 . 66 1 . 4

An ideal gas has a specific heat at constant pressure = C P = 5 R 2 . The gas is kept in a closed vessel of volume 0.0083 m 3 , at a temperature of 300 K and a pressure of 1.6 × 10 6 N m − 2 . An amount of 2 .49 × 10 4 J of heat energy is supplied to the gas. If the final temperature and pressure of the gas are 6.75 × 10 x K and 3.6 × 10 y N m - 2 respectively then find y - x .

We know that PV = nRT Number of moles n = P V R T = 1.6 × 10 6 0.0083 8.3 300 = 16 3 According to the Ist law of thermodynamics , dQ = dU + dW dQ = dU + PdV But heat is supplied at constant volume therefore dV = 0 dQ = dU = nC V dT So ∆ U = n C v Δ T = Q C p - C v = R ⇒ C v = C p - R = 3 R 2 ⇒ Q n C v = T 2 - T 1 ⇒ T 2 = T 1 + Q n C V = 300 + 2.49 × 10 4 16 3 3 R 2 ⇒ T 2 = 300 + 375 = 675 K As V = constant So P 2 P 1 = T 2 T 1 ⇒ P 2 = 675 300 1.6 × 10 6 = 3.6 × 10 6 N m − 2 On comparing, we find y=6 and x=2. Therefore y-x=4

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Graph shows a hypothetical speed distribution for a sample of $N$ gas particle:( for $V > V_{0}; \frac{d N}{d V} = 0$ )

50% studentsanswered this correctly

Important Questions on Kinetic Theory of Gases

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Physics>Chapter 18 - Kinetic Theory of Gases>Exercise-3>Q 9

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at

0 ° C

, at a pressure of

76 cm

of mercury. One of the bulbs is then placed in the melting ice, and the other is placed in a water bath maintained at

62 ° C

. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Physics>Chapter 18 - Kinetic Theory of Gases>Exercise-3>Q 10

A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle

5 cm

containing mercury and the two equal ends containing air at the same pressure

P

. When the tube is held at an angle of

60^{°}

with the vertical direction, the length of the air column above and below the mercury column are

46 cm

and

44.5 cm

respectively. Calculate the pressure

P

in centimetres of mercury. (The temperature of the system is kept at

30^{°} C

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Physics>Chapter 18 - Kinetic Theory of Gases>Exercise-3>Q 12

An oxygen cylinder of volume $30$ litres has an initial gauge pressure of $15 atm$ and a temperature of $27 ° C$ . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11 atm$ and its temperature drops to $17 ° C$ . Estimate the mass of oxygen taken out of the cylinder ( $R = 8 .31 {Jmol}^{- 1} K^{- 1}$ , molecular mass of $O_{2} = 32 u$ ).

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Physics>Chapter 18 - Kinetic Theory of Gases>Exercise-3>Q 13

Figure shows plot of

P V / T

versus

P

for

1.00 \times 10^{- 3} kg

of oxygen gas at two different temperatures.

Question Image

What is the value of $P V / T$ where the curves meet on the $y$ -axis?

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Physics>Chapter 18 - Kinetic Theory of Gases>Exercise-3>Q 15

An ideal gas having initial pressure

P

, volume

V

and temperature

T

is allowed to expand adiabatically until is volume becomes

5.66 V

while its temperature falls to

\frac{T}{2}

(i) How many degrees of freedom do gas molecules have?

(ii) Obtain the work done by the gas expansion as a function of the initial pressure $P$ and volume $V$ . [Take $[Take {(5.66)}^{0.4} = 2]$

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Physics>Chapter 18 - Kinetic Theory of Gases>Exercise-3>Q 16

An ideal gas has a specific heat at constant pressure

= C_{P} = \frac{5 R}{2}

. The gas is kept in a closed vessel of volume

0.0083 m^{3}

, at a temperature of

300 K

and a pressure of

1.6 \times 10^{6} N m^{- 2}

. An amount of

2 .49 \times 10^{4} J

of heat energy is supplied to the gas. If the final temperature and pressure of the gas are

6.75 \times 10^{x} K

and

3.6 \times 10^{y} N m^{- 2}

respectively then find

(y - x)

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Physics>Chapter 18 - Kinetic Theory of Gases>Exercise-3>Q 17

One mole of monoatomic ideal gas undergoes a process

A B C

as shown in figure. The maximum temperature of the gas during the process

A B C

is in the form

\frac{x P V}{R}

Find

x

Question Image

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Physics>Chapter 18 - Kinetic Theory of Gases>Exercise-3>Q 18

An ideal monoatomic gas occupies volume

10^{- 3} m^{3}

at temperature

3 K

and pressure

10^{3} Pa

. The internal energy of the gas is taken to be zero at this point. It undergoes the following cycle : The temperature is raised to

300 K

at constant volume, the gas is then expanded adiabatically till the temperature is

3 K

, followed by isothermal compression to the original volume. Calculate (i) The work done and the heat transferred in each process and the internal energy at the end of each process, (ii) The thermal efficiency of the cycle.

Graph shows a hypothetical speed distribution for a sample of N gas particle:( for V>V0;dNdV=0 )

Important Questions on Kinetic Theory of Gases

Learn and Prepare for any exam you want !

Graph shows a hypothetical speed distribution for a sample of $N$ gas particle:( for $V > V_{0}; \frac{d N}{d V} = 0$ )