Question

2 m3 of an ideal gas are compressed from 100 kPa to 200 kPa. As a result of the process, the internal energy of the gas increases by 10 kJ, and 150 kJ of heat is transferred to the surroundings. How much work was done by the gas during the process?

Answer 1

Answer:

work done is -150 kJ

Explanation:

given data

volume v1 = 2 m³

pressure p1 = 100 kPa

pressure p2 = 200 kPa

internal energy = 10 kJ

heat is transferred  = 150 kJ

solution

we know from 1st law of thermodynamic is

Q = du +W    ............1

put here value and we get

-140 = 10 + W

W = -150 kJ

as here work done is -ve so we can say work is being done on system

Answer 2

The work done by the gas during the process is –160 KJ

From the question given above, the following data were obtained:

• Change in internal energy (ΔU) = 10 KJ

• Heat transfered (Q) = –150 KJ

• Work (W) = ?

The work done by the gas during the process can be obtained as follow:

ΔU = Q – W

10 = –150 – W

Collect like terms

10 + 150 = –W

–W = 160 KJ

W = –160 KJ

Therefore, the work done is –160 KJ

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