In an isothermal process at 300 K, 1 mole of an ideal gas expands from a pressure 100 atm against an external pressure of 50 atm. Then total entropy change (Cal K- 1 ) in the process is -

Question

In an isothermal process at 300 K, 1 mole of an ideal gas expands from a pressure 100 atm against an external pressure of 50 atm. Then total entropy change (Cal K- 1 ) in the process is - (A) +0.39 (B) - 0.39 (C) +1.59 (D) -1.59

Tardigrade Admin · Accepted Answer

Δ Ss y s=1× 2× 2.303log(100/50)∼ eq1.39 Cal K- 1 Δ Ss u r r=(- 1 × 2 × 300 (1 - (50/100))/300)=-1.0 Cal K- 1 Δ St o t a l=+0.39calK- 1

Q. In an isothermal process at 300 K, 1 mole of an ideal gas expands from a pressure 100 atm against an external pressure of 50 atm. Then total entropy change (Cal $K^{- 1}$ ) in the process is -

+0.39

- 0.39

+1.59

-1.59

$Δ S_{sys} = 1 \times 2 \times 2.303 l o g \frac{100}{50} \sim e q 1.39 C a l K^{- 1}$

$Δ S_{s u rr} = \frac{- 1 \times 2 \times 300 ( 1 - \frac{50}{100} )}{300} = - 1.0 C a l K^{- 1}$

$Δ S_{t o t a l} = + 0.39 c a l K^{- 1}$

Q. In an isothermal process at 300 K, 1 mole of an ideal gas expands from a pressure 100 atm against an external pressure of 50 atm. Then total entropy change (Cal K−1 ) in the process is -