1 kJ of heat is given to one mole of an ideal gas and its temperature rises 0 ° C to 100 ° C . If it is kept at a constant pressure throughout the heating, find the change in its internal energy.

In the given question we have to use the concept of internal energy. The change in internal energy will be given by ∆ U = ∆ U f - ∆ U i ⇒ ∆ U = n C v ∆ T - 1000 w h e r e C v = s p e c i f i c h e a t a t c o n s tan t v o l u m e Q = h e a t T = t e m p e r a t u r e n = n o . o f m o l e s ⇒ ∆ U = 1 × 3 2 × 8 . 4 × 100 - 1000 = 1260 - 1000 = 260 J = 0 . 026 K J

One-half mole of helium gas is confined to a container at standard pressure and temperature 1 . 01 × 10 5 Pa , 273 K . How much heat is needed to double the pressure of the gas.

Here, n = 1 2 , C v = 3 J g − 1 K − 1 , M = 4 g m o l − 1 ∴ C v = M C V 4 × 3 = 12 J m o l − 1 K − 1 At constant volume P ∝ T . ∴ P 2 P 1 = T 2 T 1 = 2 , T 2 = 2 T 1 Where C v = specific heat at constant volume P = Pressure T = temperature Rise in temperature Δ T = T 2 − T 1 = 2 T 1 − T 1 = T 1 = 273 K Heat required, Δ Q = N C V Δ T = 21 × 12 × 273 = 1703 . 5 J

A sample of gas is allowed to expand quasistatically and adiabatically, its pressure drops from 120 to 100 Kpa , and its temperature drops from 300 K to 280 K . Is the gas monoatomic or diatomic ?

We know that for adiabatic process, T 1 γ P 1 1 - γ = T 2 γ P 2 1 - γ W h e r e T = t e m p e r a t u r e P = p r e s s u r e ⇒ T 1 γ P 1 1 - γ = T 2 γ P 2 1 - γ ⇒ 300 γ 120 1 - γ = 280 γ 100 1 - γ ⇒ γ = log 1 . 12 1 . 2 ⇒ γ = 2 ⇒ 1 + 2 f = 2 ⇒ f = 2 Hence, the gas is diatomic.

MEDIUM

11th West Bengal Board

IMPORTANT

Earn 100

Normal atmospheric pressure is $1.013 \times 10^{5} {Nm}^{- 2}$ and the latent heat of melting of ice is $80 cal g^{- 1}$ . At $0 ° C$ and normal atmospheric pressure the volume of $1 g$ of water increases from $1 {cm}^{3}$ to $1.091 {cm}^{3}$ on freezing. What will be the change in its internal energy ?

Important Questions on Thermodynamics

EASY

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 19 - Thermodynamics>EXERCISE>Q 14

1 kJ

of heat is given to one mole of an ideal gas and its temperature rises

0 ° C

100 ° C

. If it is kept at a constant pressure throughout the heating, find the change in its internal energy.

EASY

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 19 - Thermodynamics>EXERCISE>Q 15

One-half mole of helium gas is confined to a container at standard pressure and temperature

(1.01 \times 10^{5} Pa, 273 K)

. How much heat is needed to double the pressure of the gas.

EASY

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 19 - Thermodynamics>EXERCISE>Q 16

A cylinder fitted with a piston contains

0.10 mol

of air at room temperature

(20 ° C)

. The piston is pushed so slowly that the air within the cylinder remains essentially in thermal equilibrium with the surroundings. Find the work done by the air within the cylinder if the final volume is one-half the initial volume.

EASY

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 19 - Thermodynamics>EXERCISE>Q 17

A sample of gas is allowed to expand quasistatically and adiabatically, its pressure drops from

120

100 Kpa

, and its temperature drops from

300 K

280 K

. Is the gas monoatomic or diatomic ?

EASY

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 19 - Thermodynamics>EXERCISE>Q 18

Find the work done and the change is internal energy for a

6 cm

cube of iron as it is heated from

20 ° C

300 ° C

. For iron specific heat is

0.11 cal / gm ° C

and volume co-efficient of thermal expansion is

3.6 \times 10^{- 5 \circ} C^{- 1}

. The mass of the cube is

1700 g

and atmospheric pressure is

1 \times 10^{5} Pa

EASY

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 19 - Thermodynamics>EXERCISE>Q 19

A cubic metre of helium originally at

0 ° C

and

1 - atm

pressure is cooled at constant pressure until the volume is

0.75 m^{3}

. How much heat was removed ? [Take

1 - atm

pressure

1.013 \times 10^{5} Pa

and

C_{v} = 12.47 J {mol}^{- 1} K^{- 1}

]

EASY

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 19 - Thermodynamics>EXERCISE>Q 20

A carnot engine operates between

317 ° C

and

67 ° C

. What is its efficiency ?

MEDIUM

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 19 - Thermodynamics>EXERCISE>Q 21

Compare the actual efficiency with the theoretical maximum efficiency. A carnot-type engine is designed to operate between

480

and

300 K

. Assuming that the engine actually produces

1.2 kJ

of mechanical energy per kilocalorie of heat absorbed.

Normal atmospheric pressure is 1.013×105 Nm-2 and the latent heat of melting of ice is 80 cal g-1. At 0°C and normal atmospheric pressure the volume of 1 g of water increases from 1 cm3 to 1.091 cm3 on freezing. What will be the change in its internal energy ?

Important Questions on Thermodynamics

Learn and Prepare for any exam you want !