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1st Law of Thermodynamics

Two moles of ...

Question

Two moles of an ideal gas are undergone a cyclic process 1-2-3-1. If net heat exchange in the process is 300 J, the work done by the gas in the process 2-3 is

A

- 500 J

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B

- 5000 J

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C

- 3000 J

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D

None of these

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Solution

The correct option is D None of these

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Q. Consider a cyclic process ABCA performed by a sample of 2 moles of an ideal gas as shown in figure. If the net heat transfer during the cyclic process is -1200 J, the work done during the process BC
(Take R=8.314 J\mol-K)

Q. Consider the cyclic process

A B C A

2.0 mol

of an ideal gas as shown in figure. The temperature of the gas at

(A)

(B)

300 K

500 K

respectively. A total of

1200 J

heat is withdrawn from the sample in the process. Find the work done by the gas in process

B C

R = 8.3 J/mol -K

Q. An ideal gas is subjected to a cyclic process involving four thermodynamic states; the amounts of heat (Q) and work (W) involved in each of these states are

Q_{1} = 6000 J, Q_{2} = 5500 J, Q_{3} = - 3000 J, Q_{4} = 3500 J W_{1} = 2500 J, W_{2} = - 1000 J, W_{3} = - 1200 J, W_{4} = x J

. The ratio of the net work done by the gas to the total heat absorbed by the gas in '

η

'. The values of

x

η

respectively are:

Q. In the figure, a cyclic process ABCA of 3 moles of an ideal gas is given. The temperatures of the gas at B and C are 500 K and 1000 K respectively. If the work done on the gas in process CA is 2500 J, then find the net heat absorbed or released by an ideal gas. Take

R = 25 / 3 J / m o l - K

Q. The three processes in a thermodynamic cycle shown in the figure are : Process

1 \to 2

is isothermal; Process

2 \to 3

is isochoric (volume remains constant); Process

3 \to 1

is adiabatic. The total work done by the ideal gas in this cycle is

10 J

. The internal energy decreases by

20 J

in the isochoric process. The work done by the gas in the adiabatic process is

- 20 J

. The heat added to the system in the isothermal process is

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