The specific heat of argon at constant volume is 0.3122kj/kg k. Find the specific heat of argon at constant pressure if R = 8.314kj/kmol k. (molecular weight of argon is 39.95).
(A) 520.3
(B) 530.2
(C) 230.5
(D) 302.5
Answer
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Hint: 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 mathematical equation to get the required answer.
Formula used:
${C_P} - {C_V} = \dfrac{R}{M}$.
Complete step by step solution:
Heat is absorbed by the 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 $\left( {{C_V}} \right)$. 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 $\left( {{C_P}} \right)$ so, heat taken up 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 $R = 8.314kj/kmol k$. It is given that the molecular weight of argon is M = 39.95. So the molar specific heats of argon at constant volume and constant pressure respectively are,
$\eqalign{
& {C_V} = M{C_V}{\text{ and }}{C_P} = M{C_P} \cr
& \Rightarrow {C_P} - {C_V} = R \cr
& \Rightarrow M{C_P} - M{C_V} = R \cr
& \therefore {C_P} - {C_V} = \dfrac{R}{M} \cr} $
Putting all the given values in the above equation, we get,
$\eqalign{
& {C_P} = 0.3122 = \dfrac{{8.314}}{{39.95 \times {{10}^3}}} = 0.2081 \times {10^3} \cr
& \Rightarrow {C_P} = 0.2081 \times {10^3} + 0.3122 \cr
& \Rightarrow {C_P} = 0.5203 \times {10^3} \cr
& \therefore {C_P} = 520.3 \cr} $
So, the specific heat of argon at constant pressure is 520.3.
Hence the correct option is (a).
Note:
It is proven that for any substance, the specific heat at the constant pressure is greater than 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.
Formula used:
${C_P} - {C_V} = \dfrac{R}{M}$.
Complete step by step solution:
Heat is absorbed by the 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 $\left( {{C_V}} \right)$. 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 $\left( {{C_P}} \right)$ so, heat taken up 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 $R = 8.314kj/kmol k$. It is given that the molecular weight of argon is M = 39.95. So the molar specific heats of argon at constant volume and constant pressure respectively are,
$\eqalign{
& {C_V} = M{C_V}{\text{ and }}{C_P} = M{C_P} \cr
& \Rightarrow {C_P} - {C_V} = R \cr
& \Rightarrow M{C_P} - M{C_V} = R \cr
& \therefore {C_P} - {C_V} = \dfrac{R}{M} \cr} $
Putting all the given values in the above equation, we get,
$\eqalign{
& {C_P} = 0.3122 = \dfrac{{8.314}}{{39.95 \times {{10}^3}}} = 0.2081 \times {10^3} \cr
& \Rightarrow {C_P} = 0.2081 \times {10^3} + 0.3122 \cr
& \Rightarrow {C_P} = 0.5203 \times {10^3} \cr
& \therefore {C_P} = 520.3 \cr} $
So, the specific heat of argon at constant pressure is 520.3.
Hence the correct option is (a).
Note:
It is proven that for any substance, the specific heat at the constant pressure is greater than 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.
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