If 2 mole of an ideal gas expanded isothermally at 300 K from 5 litre volume to 50 litre volume then maximum work done in this process will be

Question

If 2 mole of an ideal gas expanded isothermally at 300 K from 5 litre volume to 50 litre volume then maximum work done in this process will be (A) - 11.49 kJ (B) +11.49 kJ (C) 4.99 kJ (D) - 4.99 kJ

Tardigrade Admin · Accepted Answer

During expansion, max. work done will be in reversible process Hence, w = -2.303 nRT log ((V2/V1)) = -2.303 × 2 × 8.314 × 300 log ((50/5)) = -11.49 kJ

Q. If 2 mole of an ideal gas expanded isothermally at 300 K from 5 litre volume to 50 litre volume then maximum work done in this process will be

- 11.49 kJ

+11.49 kJ

4.99 kJ

- 4.99 kJ

During expansion, max. work done will be in reversible process
Hence, w = -2.303 nRT log $(\frac{V _{2}}{V _{1}})$
= $- 2.303 \times 2 \times 8.314 \times 300 l o g$ $(\frac{50}{5})$
= -11.49 kJ

Q. If 2 mole of an ideal gas expanded isothermally at 300 K from 5 litre volume to 50 litre volume then maximum work done in this process will be

- 11.49 kJ

+11.49 kJ

4.99 kJ

- 4.99 kJ

During expansion, max. work done will be in reversible process Hence, w = -2.303 nRT log (V1​V2​​) = −2.303×2×8.314×300log (550​) = -11.49 kJ

During expansion, max. work done will be in reversible process
Hence, w = -2.303 nRT log $(\frac{V _{2}}{V _{1}})$
= $- 2.303 \times 2 \times 8.314 \times 300 l o g$ $(\frac{50}{5})$
= -11.49 kJ