Question

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12) In the P-V diagram shown in the figure, an ideal gas is processed to extract work. (R-8314 J/mol K, 1m3=1000L, 1atm=105Pa) The process from a to b is isothermal. a) Calculate work when isothermally gas is expanded from 2 L to 10 L b) Calculate work when the gas is compressed from 10 L to 2 L c) Calculate heat that needs to be added from c to a? d) What is Pa? If there is 1 mole gas, what is internal energy at a and b? Pa | | で- 2L 10L

Answer 1

12 a) The work done in isothermal expansion from 2L to 10L is given by :

$W_{1}=\int_{V_{a}}^{V_{b}}PdV=\int_{V_{a}}^{V_{b}}(\mu RT/V) dV=\mu RTln(V_{b}/V_{a})$ since the equation of state for an ideal gas is $PV=\mu RT$.

We have $\mu RT=P_{a}V_{a}=P_{b}V_{b}=1atm*10L=10^{5}N/m^{2}*10*10^{-3}m^{3}=10^{3}J.$

Therefore $W_{1}=10^{3}J*ln(10L/2L)=10^{3}ln5 J=1600 J.$ .

b)The work done in compressing the gas from 10L to 2L in process 2 is given by:

$W_{2}=\int_{V_{b}}^{V_{c}}PdV=P\int_{V_{b}}^{V_{c}}dV=1 atm*(V_{c}-V_{b})=10^{5}N/m^{2}*(2*10^{-3}-10*10^{-3})m^{3}=-8*10^{2}J=-800 J.$ since the pressure is constant for this process.

c)For the whole process from the first law of thermodynamics we have:

$\Delta U=\Delta Q-\Delta W=0$. i.e the change in internal energy of the gas is 0.Therefore $\Delta Q=\Delta W=W_{1}+W_{2}=1600-800 J=800 J.$

This is the heat that needs to be added from c to a.

d) For isothermal expansion $P_{a}V_{a}=P_{b}V_{b}$ i.e $P_{a}*2L=1 atm*10L$ or $P_{a}=5 atm$.

The expression for internal energy of the gas is $U=3/2\mu RT$. But we have $PV=\mu RT$. Therefore the internal energy of the gas at a $U_{a}=3/2P_{a}V_{a}=3/2*5 atm*2L=3/2*5*10^{5}N/m^{2}*2*10^{-3}m^{3}=1500 J.$

Similarly the internal energy of the gas at b=$U_{b}=3/2P_{b}V_{b}=3/2*1 atm*10L=3/2*1*10^{5}N/m^{2}*10*10^{-3}m^{3}=1500 J.$