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Question: Specific heat of Argon at constant pressure is \[0.125{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}}\] ...

Specific heat of Argon at constant pressure is 0.125 cal g1K10.125{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}} and at constant volume is 0.075 cal g1K10.075{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}}. Calculate the density of Argon at NTP, given J=4.18x107erg cal1J = 4.18x{10^7}erg{\text{ }}ca{l^{ - 1}} and normal pressure is 1.01x106dyne cm21.01x{10^6}dyne{\text{ }}c{m^{ - 2}}.

Explanation

Solution

The cp{c_p} and cv{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 Cp=Mcp{C_p} = M{c_p} , where
Cp{C_p} is the Molar heat of the gas at the constant pressure, MM is the mass of the gas and cp{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 Cv=Mcv{C_v} = M{c_v}, where
Cv{C_v} is the Molar heat of the gas at the constant volume, MM is the mass of the gas and cv{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 cal g1K10.125{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}}and the specific heat of Argon at constant volume is 0.075 cal g1K10.075{\text{ }}cal{\text{ }}{g^{ - 1}}{K^{ - 1}}.
Therefore, we can write it as:
(cp)Ar=0.125 cal g1K1 (cv)Ar=0.075 cal g1K1  {\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:
CpCv=RJ McpMcv=RJ M(cpcv)=RJ  {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=ρ×VM=\rho\times V
Thus the above equation becomes
ρV(cpcv)=RJ (cpcv)=RJρV  \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=PVTR=\dfrac{PV}{T} from the universal gas equation stated above, we get

cpcv=PVTJρV cpcv=PTJρ {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

cpcv=PTJρ ρ=PTJ(cpcv) {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

ρ=PTJ(cpcv) ρ=1.01×106293.15×4.18×107×(0.1250.075) ρ=1.01×106293.15×4.18×107×0.05 ρ=0.001649 g cm3 \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.