Question

One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 atm. Determine the value of q, w.

(b) Now assume the gas expands by the same amount again isothermally but now irre-

versibly against 1 atm pressure (instead of reversible expansion) and calculate again q, w, delta U and delta H.

Answer 1

Step-by-step explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature
`(T_i)=300 K`

`P_i=10 atm`

`P_f=2 atm`

Work done in iso-thermal process
`=P_iV_iln(P_i)/(P_f)`

`P_i`=initial pressure

`P_f`=Final Pressure

`W=10* 2.463* ln(10)/(2)=39.64 J`

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is
`=P\Delta V`

`V_1=(RT_1)/(P_1)=(1* 0.0821* 300)/(10)=2.463 L`

`V_2=(RT_2)/(P_2)=(1* 0.0821* 300)/(2)=12.315 L`

`\Delta W=1* (12.315-2.463)=9.852 J`

`\Delta q=\Delta W=9.852 J`

`\Delta U=0`