Question

1-kg water in a frictionless piston-cylinder device is initially
at 250°C and 300 kPa (state

1). A total of 700 kJ of work is done **ON** the water
in order to isothermally reduce its volume to

1/20 of its initial volume (state 2). Determine the magnitude and
direction of the heat transfer

involved in this process.

Answer: -1147 kJ.

Answer #1

According to the question;

As we know that;

Work done in isothermal process, W = nRTln(Vf/Vi)

Number of moles in 1 kg of water, n = 1000/18 = 55.55

So,

R = 8.31 J/mol-K

T= 523 K

Vf/Vi = 1/20

W = (55.55)*(8.31)*(523)*ln[1/20] = -723.25 kJ

Work done on the system = 700 kJ

(work done on the system is taken as negative)

Since,

It is an isothermal process, change in internal energy is zero that is du = 0

According to 1st law of thermodyamics

du = dq - dw

Hence,

dq = dw

Put values

dq = -723.25 - 700 = -1423.25 kJ

Then,

Heat is transferred to the surroundings and its value is

=1423.25 kJ

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