`10g` of argon gas is compressed isothermally and reversibly at a temperature of `27^(@)C` from `10L` to `5L`. Calculate `q, w, DeltaU`, and `DeltaH` for this process. `R = 2.0 cal K^(-1) "mol"^(-1), log_(10)2 = 0.30`. Atomic weight of `Ar = 40`.

Explanation

Moles of argon = `(10)/(40)`
Work done for isothermal reversible process
`w = -2.303 nRT "log (V_(2))/(V_(1))`
T = 273 + 27 = 300K, `V_(1) = 10L,V_(2) = 5L`
`R = 2.0 cal K^(-1) "mol"^(-1)`
`therefore w = -2.303 xx (10)/(40) xx 2.0 xx 300 "log (5)/(10) = 103.991` cal
For isothermal process,
`DeltaU = 0`
Now, `DeltaU = q + w`
`0 = q + w or q = -w`
`therefore` q = -103.991 cal.