32 gram of oxygen gas at temperature 27 ° C is compressed adiabatically to 1 3 of its initial volume. Calculate the change in internal energy ( γ = 1 . 5 for oxygen).

For the process to take place adiabatically, we know that internal energy is given as, d U = n C v d T = n R ( γ − 1 ) d T . Now, P V γ = constant . So, we can say that, V γ × n R T V = constant ⇒ T V ( γ − 1 ) = constant . So, we can say, T 1 V 1 ( γ − 1 ) = T 2 V 2 ( γ − 1 ) 300 V 1 ( γ − 1 ) = T 2 V 3 ( γ − 1 ) 300 T 2 = 1 3 1 . 5 - 1 T 2 = 300 1 / 3 = 300 3 K N = number of moles = 32 32 = 1 . Now, the internal energy can be written as, d U = 1 × R ( 1 . 5 − 1 ) × 300 1 . 732 - 1 ⇒ 2 R × 300 3 − 1 = 2 × 25 3 × 300 1 . 732 − 1 = 50 × 100 × 0 . 732 J = 3660 J .

32 gram of oxygen gas at temperature 27 ° C is compressed adiabatically to 1 3 of its initial volume. Calculate the change in internal energy ( γ = 1 . 5 for oxygen).

For the process to take place adiabatically, we know that internal energy is given as, d U = n C v d T = n R ( γ − 1 ) d T . Now, P V γ = constant . So, we can say that, V γ × n R T V = constant ⇒ T V ( γ − 1 ) = constant . So, we can say, T 1 V 1 ( γ − 1 ) = T 2 V 2 ( γ − 1 ) 300 V 1 ( γ − 1 ) = T 2 V 3 ( γ − 1 ) 300 T 2 = 1 3 1 . 5 - 1 T 2 = 300 1 / 3 = 300 3 K N = number of moles = 32 32 = 1 . Now, the internal energy can be written as, d U = 1 × R ( 1 . 5 − 1 ) × 300 1 . 732 - 1 ⇒ 2 R × 300 3 − 1 = 2 × 25 3 × 300 1 . 732 − 1 = 50 × 100 × 0 . 732 J = 3660 J .

The following figure is given for 1 mole gas. Then, find out net work done in the cyclic process.

In the given figure, for the path B A , work done = n R T ln V A V B = − 200 R ln 2 where, R = 8 . 314 J mol - 1 K - 1 . For path B C , volume is constant as we can see from the graph. Work done will be = P ∆ V = 0 For the path D C , work done, = n R T ln V C V D = 400 R ln 2 Again, in the path A D , volume is constant. So, work done = 0 . Hence, the net work done = 200 R ln 2 . So, on solving, we get 1152 . 6 J .

The coefficient of performance of a carnot refrigerator is 5 3 . If the refrigerator operates at an intake temperature of - 23 ° C , what will be the exhaust temperature?

A refrigerator works in a reverse Carnot cycle. It has three main components: the source, the sink and the working substance. Its performance is analysed by the coefficient of performance which is the ratio of heat released to the sink to the work done to release the heat. Coefficient of performance for a carnot engine is given by, C O P = T low T high − T low , ⇒ 5 3 = 273 - 23 T high - 273 - 23 ⟹ T high − 250 = 150 ⟹ T high = 400 K = 127 ° C .

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$32 gram$ of oxygen gas at temperature $27 ° C$ is compressed adiabatically to $\frac{1}{3}$ of its initial volume. Calculate the change in internal energy ( $γ = 1.5$ for oxygen).

Important Questions on Thermodynamics

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Beta Question Bank for Medical: Physics>Chapter 17 - Thermodynamics>BEGINNER'S BOX - 4>Q 5

An ideal gas expands isothermally along $A B$ and does $700 J$ of work.

Question Image

(a) How much heat does the gas exchange along $A B$ .

(b) The gas expands adiabatically along $B C$ and does $400 J$ of work. When the gas returns to $A along C A$ , it exhausts $100 J$ of heat to its surroundings. How much work is done on the gas along this path.

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Beta Question Bank for Medical: Physics>Chapter 17 - Thermodynamics>BEGINNER'S BOX - 4>Q 6

For one mole of an ideal gas at $300 K$ occupies a volume of $0.36 m^{3}$ at a pressure of $2 atm$ . The gas expands adiabatically until its volume becomes $1.44 m^{3}$ . Next, the gas is compressed isobarically to its original volume. Finally, the pressure is increased isochorically until the gas returns to as initial state (see figure).

Question Image

(a) Determine the temperature at the point $B and C$ .

(b) The work done during the process $A to B$ .

(Assume, $γ = 1.5, R = 8 J {mol}^{- 1} K^{- 1}$ .)

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Beta Question Bank for Medical: Physics>Chapter 17 - Thermodynamics>BEGINNER'S BOX - 4>Q 7

The following figure is given for $1 mole$ gas.

Question Image

Then, find out net work done in the cyclic process.

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Beta Question Bank for Medical: Physics>Chapter 17 - Thermodynamics>BEGINNER'S BOX - 5>Q 1

An ideal engine which works between 27ºC & 227ºC takes 100 calorie of heat in one cycle. Calculate the work done in one cycle?

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Beta Question Bank for Medical: Physics>Chapter 17 - Thermodynamics>BEGINNER'S BOX - 5>Q 2

The coefficient of performance of a carnot refrigerator is

\frac{5}{3}

. If the refrigerator operates at an intake temperature of

- 23 ° C

, what will be the exhaust temperature?

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Beta Question Bank for Medical: Physics>Chapter 17 - Thermodynamics>BEGINNER'S BOX - 5>Q 3

A Carnot engine absorbs

1000 J

of heat from a reservoir at

127 ° C

and rejects

600 J

of heat during each cycle. Calculate the efficiency of the engine,

(i i)

temperature of the sink and,

(i i i)

amount of useful work done during each cycle.

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Beta Question Bank for Medical: Physics>Chapter 17 - Thermodynamics>BEGINNER'S BOX - 5>Q 4

Two carnot engines

A and B

use one after another. Engine

A

absorb heat from source at,

T_{1} = 800 K

and release heat to sink of temperature

T_{2} K

. Second engine

B

absorb heat which is released by first (at

T_{2} K

) and release heat at another sink at temperature,

T_{3} = 300 K

. If work done by two engine is same, calculate the value of

T_{2}

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Beta Question Bank for Medical: Physics>Chapter 17 - Thermodynamics>BEGINNER'S BOX - 5>Q 5

A carnot engine has an efficiency of

40 %

and temperature of sink is

300 K

. To increase efficiency up to

60 %

, calculate the change in temperature of source?

32 gram of oxygen gas at temperature 27°C is compressed adiabatically to 13 of its initial volume. Calculate the change in internal energy (γ=1.5 for oxygen).

Important Questions on Thermodynamics

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$32 gram$ of oxygen gas at temperature $27 ° C$ is compressed adiabatically to $\frac{1}{3}$ of its initial volume. Calculate the change in internal energy ( $γ = 1.5$ for oxygen).