Question

74. A quantity of 0.50 mole of an ideal gas at 20°C
expands isothermally against a constant pressure of
2.0 atm from 1.0 L to 5.0 L. Entropy change of the
system is
(a) 6.7 J K-1
(b) - 2.8 JK-1
(c) 3.9JK-
(d) None is correct​

Answer 1

Answer:

A)

Explanation:

Del S = n[Cv*ln(T2/T1) + RTln(v2/v1)]

As we know process is isothermal,

Hence T2=T1

Del S=  n[RTln(v2/v1)]

Put values get Answer !

Answer 2

Answer: The entropy change of the system is 6.7 JK⁻¹ i.e.option(a).

Explanation:

We will use the following relation to solve this question,

$\Delta S=2.303nRlog\frac{V_2}{V_1}$       (1)

Where,

ΔS=change in entropy of the system

n=number of moles of the gas

R=universal gas constant=8.314 Jmol⁻¹K⁻¹

V₂=final volume of the gas

V₁=initial volume of the gas

From the question we have,

n=0.50 mole

Temperature(T)=20°C

Pressure=2 atm

V₁=1L

V₂=5L

By substituting the values in equation (1) we get;

$\Delta S=2.303\times 0.5\times 8.314\times log\frac{5}{1}$

$\Delta S=6.68\approx6.7JK^-^1$

Hence, the entropy change of the system is 6.7 JK⁻¹ i.e.option(a).