Question

One gram-mole of ideal gas is contained in a piston-cylinder assembly. Cp=(7/2)R, Cv=(5/2)R. The gas expands from 3 to 1 atm. Heat of 1000J is transferred to the gas during the process. External pressure maintains at 1 atm throughout. Initial temperature of the gas is 300K. Find work and internal energy change.

Answer 1

Here, it is seen that during this process pressure is dropping from 3 atm to 1 atm and 1000J heat is added and initial temperature is 300K

Now, this process is can't be polytropic because in that process heat addition is zero.

And this process can't be isobaric and isochoric

The only process which is well suited to the given data is an isothermal process.

Here P1=3

P2= 1

Q = 1000J

T1=300K

T2= 300K

We have to find Work done

W= 2.3P1*VI*log(P1/P2)

Calculate V1

P1*V1=mRT

VI=mRT/P1

V1= 1*8.314*300/3

V1= 831.4

Now Work done

W= 2.3*3*831.4*log(3/1)

W= 1000J

Now Internal Energy

dU= m*Cv*(T2-T1)

As T2-T2=0

So, dU=0

Internal Energy is Zero