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Calculate w, q and DeltaU when 0.75 mol ...

Calculate w, q and `DeltaU` when 0.75 mol of anideal gas expands isothermally and revensibly at `27^(@)C` from a volume of 15L to 25L

Text Solution

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For isothermal reversible expansion of an ideal gas,
`w= -nRT ln (V_(2))/(V_(1))= -2.303 nRT log. (V_(2))/(V_(1))`
Putting `n= 0.75` mol, `V_(1) =15 L, V_(2) = 25 L, T= 27 + 273 = 300 K` and `R = 8.314 J K^(-1) mol^(-1)` , we get `w = -2.303 xx 0.75 xx 8.314 xx 300 log. (25)/(15) = -955.5 J` `( -` ve sign represents work of expansion)
For isothermal expansion of an ideal gas, `Delta U = 0`
`:. Delta U = q+w` gives ` q= -w =+ 955.5 J`
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Knowledge Check

  • Calculate the work done when 2 moles of hydrogen expand isothermally and reversibly at 27^(@)C from 15 to 50 litres.

    A
    `14.45Kcal`
    B
    `1445J`
    C
    `-1445cal`
    D
    `14.45KJ`

  • Calculate the work done (in joules) when 0.2 mole of an ideal gas at 300 K expands isothermally and reversibly from initial volume of 2.5 litres to the final volume of 25 litres.

    A
    996
    B
    1148
    C
    11.48
    D
    897

  • If x mole of ideal gas at 27^(@)C expands isothermally and reversibly from a volume of y to 10 y, then the work done is

    A
    `w = x R 300 In y`
    B
    `w = -300 x R In y/10y`
    C
    `w = - 300 x R ln 10`
    D
    `w = 100 x R In 1/y`