Question

5 moles of an ideal gas expands reversibly from a volume of 8 dm cube to 80 dm cube at a temperature of 27 degree Celsius calculate the change in entropy and change in internal energy

Answer 1

Answer:

$V_{1} =8cm^{3}$

$V_{2} =80dm^{3}$

n=5

T=27+273=300 K

Δ$S_{T} =2.303nRlog\frac{V_{2} }{V_{1} }$

Let us substitute the given values in the formula we get

Δ$S_{T} =2.303*5*8.314*log\frac{80}{8}$

                   =95.74JK

ΔG=2.303nRT$log\frac{V_{1} }{V_{2} }$

then substitute the values we get

ΔG=2.303*5*8.314*$log\frac{80}{8} *300$

     =-28720J

Answer 2

The change in entropy and change in internal energy are 95.72 J/K and 0 Joules.

Explanation:

a) To calculate change in internal energy for isothermal, reversible expansion

$\Delta U=nC_vdT$

$\Delta U$ = internal energy change

$C_v$ = specific heat at constant Volume

$dT$ = change in temperature = 0

Thus $\Delta U=0$

Thus the change in internal energy is 0 Joules

b) To calculate the entropy change for isothermal, reversible expansion process, we use the equation:

$\Delta S=nR\ln(\frac{V_2}{V_1})$

where,

$\Delta S$ = Entropy change

n = number of moles = 5 mole

R = Gas constant = 8.314 J/mol.K

$V_1$ = initial volume = $8dm^3$

$V_2$ = final volume = $80dm^3$

Putting values in above equation, we get:

$\Delta S=5mol\times 8.314J/mol.K\times \ln(\frac{80}{8})\\\\\Delta S=95.72J/K$

Hence, the entropy change of the process is 95.72 J/K.

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