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Specific heat of Argon at constant pressure is \[0.125{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}}\] and at constant volume is \[0.075{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}}\]. Calculate the density of Argon at NTP, given \[J = 4.18x{10^7}erg{\text{ }}ca{l^{ - 1}}\] and normal pressure is \[1.01x{10^6}dyne{\text{ }}c{m^{ - 2}}\].

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Last updated date: 25th Jun 2024
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Answer
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Hint: The ${c_p}$ and ${c_v}$ values of Argon are given in the question. Using the prerequisite knowledge that Argon is a monatomic gas, we can easily solve the above question using the standard formula substitutions.

Formula Used:
1. The molar heat of a gas at constant pressure is given by ${C_p} = M{c_p}$ , where
${C_p}$ is the Molar heat of the gas at the constant pressure, $M$ is the mass of the gas and ${c_p}$ is the specific heat of the gas at the constant pressure.
2. The molar heat of a gas at constant volume is given by ${C_v} = M{c_v}$, where
${C_v}$ is the Molar heat of the gas at the constant volume, $M$ is the mass of the gas and ${c_v}$ is the specific heat of the gas at the constant volume.

Complete step by step answer:
In the above question, it is given that the specific heat of Argon at constant pressure is \[0.125{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}}\]and the specific heat of Argon at constant volume is \[0.075{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}}\].
Therefore, we can write it as:
$
{\left( {{c_p}} \right)_{Ar}} = 0.125{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}} \\
{\left( {{c_v}} \right)_{Ar}} = 0.075{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}} \\
$
Using the formulae for the molar heats of Argon at constant pressure and volume and the difference between them, as stated above, we can establish the following relation:
$
{C_p} - {C_v} = \dfrac{R}{J} \\
\Rightarrow M{c_p} - M{c_v} = \dfrac{R}{J} \\
\Rightarrow M\left( {{c_p} - {c_v}} \right) = \dfrac{R}{J} \\
$
Now, using the formula for the density of a gas as stated above, we can substitute the following value of M in the above equation - $M=\rho\times V$
Thus the above equation becomes
$
\rho V\left( {{c_p} - {c_v}} \right) = \dfrac{R}{J} \\
 \Rightarrow \left( {{c_p} - {c_v}} \right) = \dfrac{R}{{J\rho V}} \\
$
Now, substituting the following value of R, ie.,$R=\dfrac{PV}{T}$ from the universal gas equation stated above, we get
\[
{c_p} - {c_v} = \dfrac{{PV}}{{TJ\rho V}} \\
\Rightarrow{c_p} - {c_v}= \dfrac{P}{{TJ\rho }} \\
 \]
Hence, to get the density of Argon, we solve for , as shown below
\[
{c_p} - {c_v} = \dfrac{P}{{TJ\rho }} \\
\Rightarrow\rho = \dfrac{P}{{TJ\left( {{c_p} - {c_v}} \right)}} \\
\]
Putting values of all the constants and variables in the equation, we get
\[
\rho = \dfrac{P}{{TJ\left( {{c_p} - {c_v}} \right)}} \\
\Rightarrow\rho= \dfrac{{1.01 \times {{10}^6}}}{{293.15 \times 4.18 \times {{10}^7} \times \left( {0.125 - 0.075} \right)}} \\
\Rightarrow\rho= \dfrac{{1.01 \times {{10}^6}}}{{293.15 \times 4.18 \times {{10}^7} \times 0.05}} \\
\therefore\rho= 0.001649{\text{ }}g{\text{ }}c{m^{ - 3}} \\
\]
Hence,the density of Argon is calculated to be 0.001649 g cm-3.

Note:This problem is a simple formula-based one but requires heavy calculations. While attempting such questions, students should try to quickly substitute the formulae values and then proceed with the calculations patiently, and not vice versa. Atomicity of Argon is 3 and specific heat for isothermal process is $\infty $ and adiabatic process is 0.