2 mol of an ideal gas is compressed by an isothermal reversible process at `27^(@)C`. As a result, pressure of the gas increases from 1 to 10 atm. Calculate `w, q, Delta U and Delta H` for the process.

Explanation

Work done in isothermal compression of an ideal gas,
`w= - 2.303 nRT " log " (P_(1))/(P_(2))[P_(1) lt P_(2)]`.
` therefore w= - 2.303xx2xx8.314xx300 " log " (1)/(10) J= 11.48 kJ`.
In isothermal process for an ideal gas, `Delta U=0 and Delta H= 0` . From the 1st law of thermodynamics, we know `Delta U= q+w`.
`therefore 0 = q+11.48 kJ , therefore q= -11.48 kJ`.