Question

During an isothermal process one mole of a monoatomic gas did 3000 J of work on its surrounding. The final volume and pressure of the gas are 25 L and 1 atm, respectively. What was the initial volume of the gas

Answer 1

Answer: The initial volume of the gas is 7.72 L

Step-by-step explanation:

For an isothermal process the temperature is constant.

`PV=nRT`

as P = pressure = 1 atm ,

V = Volume = 25 L

n = moles

R= gas constant

T = temperature

`PV=1atm* 25L`

`nRT=25Latm=25* 101.3J=2532.5J` (1Latm=101.3 J)

For isothermal reaction :

`w=-2.303nRT\log(V_2)/(V_1)`

where , w = work done by system = -ve

n = moles = 1

`V_2` = final volume = 25 L

`V_1` = initial volume = ?

`-3000J=-2.303* 2532.5\log (25)/(V_1)`

`V_1=7.72L`

Thus initial volume of the gas is 7.72 L