A100 w electric heater (1 w = 1 j/s) operates for 11 min to heat the gas in a cylinder. at the same time, the gas expands from 1 l to 6 l against a constant external pressure of 3.527 atm. what is the change in internal energy of the gas? (1 l·atm = 0.1013 kj)

The change in the internal energy of the system is 67.787 kJ.

The given parameters;

The heat added to the system by the heater;

Q = Pt = 100 x 660 = 66,000 J = 66 kJ

The work done on the system is calculated as follows;

W = PΔV

W = 3.527(6 - 1)

W = 17.64 L.atm

1 Latm = 0.1013 kJ

17.64 Latm = 1.787 kJ

The change in the internal energy of the system is calculated by applying the first law of thermodynamics as follows;

ΔU = Q + W

ΔU = 66 kJ  +  1.787 kJ

ΔU = 67.787 kJ.

Thus, the change in the internal energy of the system is 67.787 kJ.

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Explanation:

As we know that power of the heater is given as

P = 100 W

now the energy given by the heater for 11 min of time

now from 1st law of thermodynamics we know that

work done under constant pressure condition we have

now from first equation we have

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(ii) size of the scattering particle  

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