Question

A gas ${{C}_{v,m}}=\dfrac{5}{2}R$ behaving ideally was allowed to expand reversibly and adiabatically from 1 litre to 32 litre. It’s initial temperature was ${{327}^{0}}C$. The molar enthalpy change (in J/mol) for the process is:
A. -1125R
B. -675 R
C. -1575 R
D. none of these

Answer 1

Hint: Since it is given that the gas is expanding adiabatically between two volumes, you can use the relation $(\dfrac{{{T}_{2}}}{{{T}_{1}}})={{(\dfrac{{{V}_{1}}}{{{V}_{2}}})}^{^{\gamma }-1}}$. Also remember that
\[{{C}_{p}}={{C}_{V}}+\text{ }R\], and \[\gamma =\text{ }\dfrac{{{C}_{p}}}{{{C}_{v}}}\]

Complete answer:
In order to answer the question, we need to look at what values are given in the question and what the question is asking. As the question is saying, the gas is expanding adiabatically, it means that we require the use of formulas related to adiabatic expansion.
As we all know that in an adiabatic system, ${{(PV)}^{\gamma }}$remains constant. So, we get the relation $(\dfrac{{{T}_{2}}}{{{T}_{1}}})={{(\dfrac{{{V}_{1}}}{{{V}_{2}}})}^{^{\gamma }-1}}$ , where ${{T}_{2}}$ is the final temperature and ${{T}_{1}}$ is the initial temperature, ${{V}_{1}}$ initial volume and ${{V}_{2}}$ expanded volume and γ being the ratio of ${{C}_{p}}$ and ${{C}_{v}}$ also known as adiabatic index.
Also remember that ${{C}_{p}}$ is the heat capacity at constant pressure, whereas ${{C}_{v}}$ is the heat capacity at constant volume.
Now,
Using $(\dfrac{{{T}_{2}}}{{{T}_{1}}})={{(\dfrac{{{V}_{1}}}{{{V}_{2}}})}^{^{\gamma }-1}}$, we have
\[\begin{array}{*{35}{l}}
 \Rightarrow ({{T}_{2}}\div 600)={{(1\div 32)}^{\dfrac{2}{5}}}. \\
   ~ \\
\end{array}\]
On solving, we get T2= 150 K
We also know that \[{{C}_{p}}={{C}_{V}}+\text{ }R\] , so
And \[\gamma =\text{ }\dfrac{{{C}_{p}}}{{{C}_{v}}}=\dfrac{\left( {{C}_{v}}+R \right)}{{{C}_{v}}}\],\[\gamma =\text{ }{{C}_{p}}\div {{C}_{v}}=\text{ }\left( {{C}_{v}}+R \right)\div {{C}_{v}}\] , which comes out to be $\dfrac{7}{5}$
From here, we can calculate the molar enthalpy change,
\[\mathbf{\Delta }{{\mathbf{H}}_{\mathbf{m}}}_{{}}=\text{ }n\text{ }{{C}_{p}}\mathbf{\Delta T}\], where \[\mathbf{\Delta }{{\mathbf{H}}_{\mathbf{m}}}\] is the molar enthalpy change, n= number of moles=1
Which is
\[\dfrac{7}{2}R\times (150-600)\]
$= -1575 R$

Hence, the molar enthalpy change( in $Jmo{{l}^{-1}}$) for the process is -1575 R.

Note:
Always remember that when the molar enthalpy change is positive, it means that heat is gained from the surroundings, which means the reaction is endothermic. If the sign is negative then it is an exothermic reaction as heat is given to its surroundings.