Question

The entropy change when two moles of ideal monoatomic gas is heated from ${200}^{\circ }C$ to ${300}^{\circ }C$ reversibly and isochorically?

• A. $\frac{3}{2}R\ln \left(\frac{300}{200}\right)$
• B. $\frac{5}{2}R\ln \left(\frac{573}{273}\right)$
• C. $3R\ln \left(\frac{573}{473}\right)$
• D. $\frac{3}{2}R\ln \left(\frac{573}{473}\right)$

Answer 1

The correct option is C $3R\ln \left(\frac{573}{473}\right)$

Entropy is given as:
$△S=n{C}_{v}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$

So, entropy change for an isochoric process (constant volume) is given as:

$△S=n{C}_{v}ln\frac{T_2}{T_1}$
where
$T_2=573K$
$T_1=473K$
$n=2$
for ideal monoatomic gas we know
$C_v=\frac{3}{2}R$

Putting the values we get:
$△S=2\times \frac{3}{2}R\times \ln \frac{573}{473}=3R\ln \left(\frac{573}{473}\right)$