Question

An ideal gas expands at a constant total pressure of 3.0 atm from 400 mL to 600 mL. heat then flows out of the gas at a constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. Calculate: (a) the total work done by the gas in the process.(b) the total heat flow into the gas.

Answer 1

Answer:

60.795 J

60.795 J

Explanation:

P = Pressure = 3 atm

$dU$ = Internal energy = $nC_vdT=0$

$V_f$ = Final volume = 600 mL

$V_i$ = Initial volume = 400 mL

Work done is given by

$dW=PdV\\\Rightarrow dW=P(V_f-V_i)\\\Rightarrow dW=3\times 101325\times (600-400)\times 10^{-6}\\\Rightarrow dW=60.795\ J$

The total work done by the gas in the process is 60.795 J

Total heat flow is given by

$dQ=dU+dW\\\Rightarrow dQ=nC_vdT+dW\\\Rightarrow dQ=0+60.795 J\\\Rightarrow dQ=60.795\ J$

The total heat flow into the gas is 60.795 J