Question

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

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

Answer 1

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

We know that for a reversible process and an ideal system $△S=n{C}_{v}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$
For an isochoric process, volume is constant i.e. $V_1=V_2$
so, $△S=n{C}_{v}ln\frac{T_2}{T_1}$
Given, number of moles, n=2
since gas is monoatomic $C_v=\frac{3R}{2}$ $T_1=473K$, $T_2=573K$
putting above values we get,
$△S=2\times \frac{3}{2}Rln\left(\frac{573}{473}\right)$
$△S=3Rln\left(\frac{573}{473}\right)$