2 mole of an ideal gas at 27 degree celcius and 1 atm pressure expand isothermally and reversibly till its pressure becomes 0.25 atm . calculate q , w and delta u

In a reversible isothermal expansion of a gas,at constant temperature,
       
T = 273 + 27 = 300K and n = 2

                                $$\begin{gathered} W=-2.303 nRT \log \frac{p_1}{p_2} \\ = -2.303 \times 2 \times 0.082 \times 300 \times \log \frac{1}{0.25} \\ = -68.21J \end{gathered}$$  

At constant temperature,
                 ΔU = 0         (First law of thermodyanamics, ΔU = q + W,and for irreversible process q=-W)
      and,
                W= -q = 68.21J