Question

An ideal gas expands at a constant total pressure of 3.0 atm from 450 mL to 850 mL . Heat then flows out of the gas at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value.

Part A Calculate the total work done by the gas in the process. Express your answer to two significant figures and include the appropriate units.

Part B Calculate the total heat flow into the gas. Express your answer to two significant figures and include the appropriate units.

Answer 1

Part A.

Work-dine by gas in the process, when pressure remains constant is given by:

W = -P*dV

Given that:

P = 3.0 atm = 3.0*1.013*10^5 Pa

Vf = 850 mL = 850*10^-6 m^3

Vi = 450 mL = 450*10^-6 m^3

So,

W = P*(Vf - Vi)

W = 3.0*1.013*10^5*(850*10^-6 - 450*10^-6)

W = 121.56 J = 1.2*10^2 J

Part B.

Using 1st law of thermodynamics:

dQ = dU + W

Since initial and final temperature are same, So dU = Change in internal energy = 0 J

So total heat flow in the gas will be:

dQ = 0 + (121.56) = 121.56 J

dQ = 1.2*10^2 J

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