Question

One mole of a monoatomic ideal gas is heated at constant pressure from 25${}^o$C to 300${}^o$C. Calculate the $\Delta$H, $\Delta$U, work done and entropy change during the process.
$(Given{C}_v=\frac{3}{2}R)$

• A. $\Delta H=1375cal,\Delta U=825cal,\Delta W=-550cal,\Delta S=3.27cal{K}^{-1}mo{l}^{-1}$
• B. $\Delta H=1325cal,\Delta U=820cal,\Delta W=-550cal,\Delta S=3.28cal{K}^{-1}mo{l}^{-1}$
• C. $\Delta H=1370cal,\Delta U=821cal,\Delta W=-550cal,\Delta S=3.27cal{K}^{-1}mo{l}^{-1}$
• D. $\Delta H=1345cal,\Delta U=841cal,\Delta W=-560cal,\Delta S=3.28cal{K}^{-1}mo{l}^{-1}$

Answer 1

The correct option is A $\Delta H=1375cal,\Delta U=825cal,\Delta W=-550cal,\Delta S=3.27cal{K}^{-1}mo{l}^{-1}$

For a monoatomic gas, $C_P-C_V=R$
$C_V=3calmo{l}^{(}-1{)}^0{C}^{-1}$ and $R=2calmo{l}^{(}-1{)}^0{C}^{-1}$
Substitute values in the above expression
$C_P-3calmo{l}^{(}-1{)}^0{C}^{-1}=2calmo{l}^{(}-1{)}^0{C}^{-1}$ or $C_P=5calmo{l}^{(}-1{)}^0{C}^{-1}$
$C_p=C_v+R=\frac{3}{2}R+R=\frac{5}{2}R$
At constant pressure, heat given is equal to enthalpy change.
$q_p=\Delta H=n\times C_p\times \Delta T=1\times \frac{5}{2}R\times [573-298]=1375cal$
The work done during the process is given by the expression $w=-nR[T_2-T_1]$
Substitute values in the above expression.
$w=-1\times 2\times [573-198]=-550cal$
The change in internal energy is $\Delta U=q+w=1375-550=825cal$
The expression for the entropy change is $\Delta S=2.303\times n\times C_{p}lo{g}_{10}\frac{T_2}{T_1}$
Substitute values in the above expression.
$\Delta S=2.303\times 1\times \frac{5}{2}\times Rlo{g}_{10}\frac{573}{298}=2.303\times \frac{5}{2}\times 2\times lo{g}_{10}\frac{573}{298}=3.27cal{K}^{-1}mo{l}^{-1}$