The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial stat (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 stat (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

In an isothermal process, internal energy remains same as temperature is constant. The change in entropy of an ideal gas

Δ S = \frac{Δ Q}{T}

...(i)
In isothermal process, temperature does not change, that is, internal energy which is a function of temperature will remain same, i.e.,

Δ U = 0.

First law of thermodynamics gives

Δ U = Δ Q - W

or

0 = Δ Q - W

or

Δ Q = W

i.e.,

Δ Q =

work done by gas in isothermal process which went through from

(P_{1}, V_{1}, T)

to

(P_{2}, V_{2}, T)

or

Δ Q = μ RT lo g_{e} (\frac{V _{2}}{V _{1}})

...(ii)
For 1 mole of an ideal gas,

μ = 1,

so from Eqs. (i) and (ii)

Δ S = R lo g_{e} (\frac{V _{2}}{V _{1}}) = R l n (\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 stat $(P_{1}, V_{1}, T)$ to the final state $(P_{2}, V_{2}, T)$ is equal to:

zero

$R l n T$

$R l n \frac{V _{1}}{V _{2}}$

$R l n \frac{V _{2}}{V _{1}}$

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

$R l n T$

$R l n \frac{V _{1}}{V _{2}}$

$R l n \frac{V _{2}}{V _{1}}$