A piston filled with `0.04` mol of an ideal gas expands reversibly from `50.0 mL` to `375 mL` at a constant temperature of `37.0^(@)C`. As it does so, it absorbs `208 J` of heat. The value of `q` and `w` for the process will be:
`(R=8.314J//molK)(ln 7.5=2.01)`

Options

A. q = +208J,w = − 208 J
B. q = -208J,w = − 208 J
C. q = -208J,w = + 208 J
D. q = +208J,w = + 208 J

Answer

A

Explanation

Method I :
The process is isothermal expansion Hence, `q = - w`
`Deltau = 0`
`q = + 208 J`
`w = -208 J` (expansion work).
Method II :
`q = + 208 J` (as it absorb heat)
`w_(rev) = - 2.303 nRT log_10(v_2/v_1)`
` - 2.303 xx 0.04 xx8.314 xx 310 log_10"" ((375)/10) = -208 J`.