Question

when one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice its initial temperature, the change in entropy (delta s) is

Answer 1

Answer : The change in entropy is $(C_v-R)\ln 2]$

Explanation :

Formula used :

$\Delta S=nC_v\ln (\frac{T_2}{T_1})+nR\ln (\frac{V_2}{V_1})$

where,

$\Delta S$ = change in entropy

n = number of moles of gas  = 1 mole

R = gas constant

$V_1$ = let the initial volume of gas  = $V$

$V_2$ = final volume of gas  = $\frac{1}{2}V$

$T_1$ = let the initial volume of gas  = $T$

$T_2$ = final volume of gas  = $2T$

Now put all the given values in the above formula, we get:

$\Delta S=(1mole)\times C_v\ln (\frac{2T}{T})+(1mole)\times R\ln (\frac{(\frac{1}{2}V)}{V})$

$\Delta S=C_v\ln (2)+R\ln (\frac{1}{2})$

$\Delta S=C_v\ln (2)+R[\ln 1-\ln 2]$

$\Delta S=C_v\ln (2)+R[0-\ln 2]$

$\Delta S=C_v\ln (2)-R\ln 2]$

$\Delta S=(C_v-R)\ln 2]$

Therefore, the change in entropy is $(C_v-R)\ln 2]$

Answer 2

Answer:

Explanation:

ΔS=nCv  ×ln(T2 /T1 )+R×ln(V2 /V21)

ΔS=Cv×ln2+R×ln 1/2

​ ΔS= Cv×ln2+R×[(ln1)-(ln2)]   (ln1=0)

​ ΔS= Cv×ln2-R×(ln2)

ΔS=(Cv −R)ln2