Question

The specific heat at argon at constant volume is $0.3122J/kg{\text{ }}K$. Find the specific heat of argon at constant pressure if $R = 8.314kJ/kmol{\text{ }}K$ (molecular weight of argon = $39.95$) in $J/kg{\text{ }}K$.

Answer 1

The difference between the two molar specific heats of an ideal gas is ${c_p} - {c_v}$. It is given that the molecular weight of argon is $M = 39.95$ and $R$ is the universal gas constant. Put all the values in the equation, ${c_p} - {c_v} = \dfrac{R}{M}$.

Complete step by step answer:

Heat absorbed by unit mass of a gas to raise its temperature by one degree, keeping the volume constant, is called the specific heat of that gas at constant volume (${c_v}$).
So, heat taken by a gas of mass $m$ for rise in temperature $t$, at constant volume, is
${Q_v} = m{c_v}t$
Again, heat absorbed by unit mass of $1$ mole of a gas to raise its temperature by one degree, keeping the pressure constant, is called the specific heat of that gas at constant pressure ( ${C_p}$ ).
So, heat taken by a gas of mass $m$ for rise in temperature $t$, at constant pressure, is
${Q_p} = m{c_p}t$
Now, the difference between the two molar specific heats of an ideal gas is ${c_p} - {c_v}$.
$R$ is the universal gas constant or molar gas constant, whose value is $R = 8.314kJ/kmol{\text{ }}K$
It is given that the molecular weight of argon is $M = 39.95$.
So, molar specific heats of argon at constant volume and constant pressure respectively, are
${C_v} = M{c_v}$ and ${C_p} = M{c_p}$
Now, ${C_p} - {C_v} = R$
or, $M{c_p} - M{c_v} = R$
or, ${c_p} - {c_v} = \dfrac{R}{M}$
Putting all the given values in the above equation, we get,
${c_p} - 0.3122 = \dfrac{{8.314}}{{39.95 \times {{10}^{ - 3}}}} = 0.2081 \times {10^3}$
or, ${c_p} = 0.2081 \times {10^3} + 0.3122 = 0.5203 \times {10^3} = 520.3$
So, the specific heat of argon at constant pressure is $520.3$.

Note:
It is proven that for any substance, the specific heat at constant pressure is greater than the specific heat at constant volume. This is because at constant volume, no work is done. So, the heat absorbed changes the internal energy only. But at constant pressure, the heat absorbed changes the internal energy and also does the same work. Thus, a greater amount of heat is absorbed in the later case, and as a result, the specific heat at constant pressure is higher.