For an isothermal reversible expansion process, the value of q can be calculated by the expression:

Question

For an isothermal reversible expansion process, the value of q can be calculated by the expression: (A) q =2.303 nRT log ( V 2/ V 1) (B) q =-23 ⋅ 3032 n RT log ( V 2/ V 1) (C) q=-2.303 n R T log (v1/v2) (D) q=-P exp n R T log (v1/v2)

Tardigrade Admin · Accepted Answer

For an isothermal reversible process, Δ T =0 ∴ Δ U = nC V Δ T =0 beginarrayl text Also, Δ U = q + w =0 ∴ q =- W endarray For an isothermal reversible expansion, w=-n R T ln ((V2/V1))= -2.303 nRT log (( V 2/ V 1)) ∴ q =- w =2.303 nRT log (( V 2/ V 1))

Q. For an isothermal reversible expansion process, the value of $q$ can be calculated by the expression:

$q = 2.303 n RT lo g \frac{V _{2}}{V _{1}}$

$q = - 23 \cdot 3032 n RT lo g \frac{V _{2}}{V _{1}}$

$q = - 2.303 n RT lo g \frac{v _{1}}{v _{2}}$

$q = - P_{e x p} n RT lo g \frac{v _{1}}{v _{2}}$

Q. For an isothermal reversible expansion process, the value of q can be calculated by the expression:

q=2.303nRTlogV1​V2​​

q=−23⋅3032nRTlogV1​V2​​

q=−2.303nRTlogv2​v1​​

q=−Pexp​nRTlogv2​v1​​

For an isothermal reversible process, ΔT=0 <br/>∴ΔU=nCV​ΔT=0<br/> <br/><br/> Also, ΔU=q+w=0<br/>∴q=−W<br/>​<br/> For an isothermal reversible expansion, w=−nRTln(V1​V2​​)= −2.303nRTlog(V1​V2​​) ∴q=−w=2.303nRTlog(V1​V2​​)

Q. For an isothermal reversible expansion process, the value of $q$ can be calculated by the expression:

$q = 2.303 n RT lo g \frac{V _{2}}{V _{1}}$

$q = - 23 \cdot 3032 n RT lo g \frac{V _{2}}{V _{1}}$

$q = - 2.303 n RT lo g \frac{v _{1}}{v _{2}}$

$q = - P_{e x p} n RT lo g \frac{v _{1}}{v _{2}}$