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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

Respuesta :

Answer:  The initial volume of the gas is 7.72 L

Explanation:

For an isothermal process the temperature is constant.  

[tex]PV=nRT[/tex]

as P = pressure = 1 atm ,

V = Volume =  25 L

n = moles

R= gas constant

T = temperature

[tex]PV=1atm\times 25L[/tex]

[tex]nRT=25Latm=25\times 101.3J=2532.5J[/tex]     (1Latm=101.3 J)

For isothermal reaction :

[tex]w=-2.303nRT\log\frac{V_2}{V_1}[/tex]

where , w = work done by  system = -ve

n = moles = 1

[tex]V_2[/tex] = final volume = 25 L

[tex]V_1[/tex] = initial volume = ?

[tex]-3000J=-2.303\times 2532.5\log \frac{25}{V_1}[/tex]

[tex]V_1=7.72L[/tex]

Thus initial volume of the gas is 7.72 L

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