The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state (P1,V1,T) to the final state (P2,V2,T) is equal to:

Question

The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state (P1,V1,T) to the final state (P2,V2,T) is equal to: (A) zero (B) R ln T (C) R ln (V1/V2) (D)  R ln (V2/V1)

Tardigrade Admin · Accepted Answer

Key Idea: In an isothermal process, there is no change in internal energy of gas, ie., Δ U=0. The change in entropy of an ideal gas Δ S=(Δ Q/T) ldots(i) In isothermal process, there is no change in internal energy of gas, i.e., Δ U=0 ∴ Δ U=Δ Q-W ⇒ 0 = Δ Q-W ⇒ Δ Q = W i.e., Δ Q= work done by gas in isothermal process which went through from (P1, V1, T) to (P2, V2, T) or Δ Q=μ R T log e((V2/V1)) ldots(ii). For 1 mole of an ideal gas, μ=1 So, from Eqs. (i) and (ii), we get or Δ S =R log e((V2/V1)) =R ln ((V2/V1))

Q. The change in the entropy of a $1$ mole of an ideal gas which went through an isothermal process from an initial state $(P_{1}, V_{1,} T)$ to the final state $(P_{2}, V_{2,} T)$ is equal to:

zero

$R ln T$

$R ln \frac{V _{1}}{V _{2}}$

$R ln \frac{V _{2}}{V _{1}}$

Q. The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state (P1​,V1,​T) to the final state (P2​,V2,​T) is equal to:

zero

RlnT

RlnV2​V1​​

RlnV1​V2​​

Q. The change in the entropy of a $1$ mole of an ideal gas which went through an isothermal process from an initial state $(P_{1}, V_{1,} T)$ to the final state $(P_{2}, V_{2,} T)$ is equal to:

$R ln T$

$R ln \frac{V _{1}}{V _{2}}$

$R ln \frac{V _{2}}{V _{1}}$