Question

The heat required to raise the temperature from 300.0 K to 400.0 K for 1 mole of a gas at constant volume is 2079 J. The internal energy required to heat the same gas at constant pressure from 550.0 K to 600.0 K is 1305 J. The gas does 150 J of work during this expansion at constant pressure. Is this gas behaving ideally? Is the gas a monatomic gas? Explain.

Short answer

The given gas is neither ideal nor monoatomic.

Step-by-step solution

Determination of behaviour of gas

Let us take heat required to raise the temperature at constant volume be Q_v and heat required to raise the temperature at constant pressure be Q_p.

Heat capacity can be written as C_v at constant volume and C_p at constant pressure.

C_v can be calculated by using the following formula,

$$\begin{gathered} {\mathit{Q}}_{\mathbf{v}}\mathbf{=}\mathit{n}{\mathit{C}}_{\mathbf{V}}\mathbf{∆}\mathit{T} \\ {\mathit{C}}_{\mathbf{v}}\mathbf{=}\frac{{\mathbf{Q}}_{\mathbf{v}}}{\mathbf{n}\mathbf{∆}\mathbf{T}} \\ {\mathit{C}}_{\mathbf{v}}\mathbf{=}\frac{\mathbf{2079}\mathbf{J}}{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{)}\mathbf{(}\mathbf{400}\mathbf{.}\mathbf{0}\mathbf{K}\mathbf{-}\mathbf{300}\mathbf{.}\mathbf{0}\mathbf{K}\mathbf{)}} \\ {\mathit{C}}_{\mathbf{v}}\mathbf{=}\frac{\mathbf{2079}\mathbf{J}}{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{)}\mathbf{(}\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{K}\mathbf{)}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{79}\mathit{J}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}{\mathit{K}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

Q_p can be calculated by using the following formula,

$\begin{array}{rcl}\mathbf{∆}\mathit{U}& \mathbf{=}& \mathit{Q}\mathbf{+}\mathit{W}\\ {\mathit{Q}}_{\mathbf{P}}& \mathbf{=}& \mathbf{∆}\mathit{U}\mathbf{-}\mathit{W}\\ {\mathit{Q}}_{\mathbf{P}}& \mathbf{=}& \mathbf{1305}\mathit{J}\mathbf{-}\mathbf{150}\mathit{J}\\ {\mathit{Q}}_{\mathbf{P}}& \mathbf{=}& \mathbf{1155}\mathit{J}\end{array}$

Now, C_p can be calculated by using the following formula,$\begin{array}{rcl}{\mathit{Q}}_{\mathbf{P}}& \mathbf{=}& \mathit{n}{\mathit{C}}_{\mathbf{P}}\mathbf{∆}\mathit{T}\\ {\mathit{C}}_{\mathbf{P}}& \mathbf{=}& \frac{{\mathbf{Q}}_{\mathbf{P}}}{\mathbf{n}\mathbf{∆}\mathbf{T}}\\ {\mathit{C}}_{\mathbf{P}}& \mathbf{=}& \frac{\mathbf{1155}\mathbf{J}}{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{)}\mathbf{(}\mathbf{600}\mathbf{.}\mathbf{0}\mathbf{K}\mathbf{-}\mathbf{550}\mathbf{.}\mathbf{0}\mathbf{K}\mathbf{)}}\\ {\mathit{C}}_{\mathbf{P}}& \mathbf{=}& \frac{\mathbf{1155}\mathbf{J}}{\mathbf{\left(}\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{\right)}\mathbf{(}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{k}\mathbf{)}}\mathbf{=}\mathbf{23}\mathbf{.}\mathbf{1}\mathit{J}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}{\mathit{K}}^{\mathbf{-}\mathbf{1}}\end{array}$

For ideal gas, C_p- C_v=R

Where, R = gas constant = 8.314 Jmol^-1K^-1

${\mathit{C}}_{\mathbf{P}}\mathbf{-}{\mathit{C}}_{\mathbf{v}}\mathbf{=}\mathbf{23}\mathbf{.}\mathbf{1}\mathit{J}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}{\mathbf{K}}^{\mathbf{-}\mathbf{1}}}\mathbf{-}\mathbf{20}\mathbf{.}\mathbf{79}\mathit{J}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}{\mathit{K}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{31}\mathit{J}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}{\mathit{K}}^{\mathbf{-}\mathbf{1}}$

Since, the value is not equal to the value of R. Thus, the given gas is not ideal.

Determination of nature of gas

For gas to be monatomic,

$$\begin{gathered} {\mathit{C}}_{\mathbf{v}}\mathbf{=}\frac{\mathbf{3}\mathbf{R}}{\mathbf{2}} \\ {\mathit{C}}_{\mathbf{P}}\mathbf{=}\mathit{R}\mathbf{+}{\mathit{C}}_{\mathbf{v}}\mathbf{+}\frac{\mathbf{5}\mathbf{R}}{\mathbf{2}} \\ \frac{{\mathbf{C}}_{\mathbf{P}}}{{\mathbf{C}}_{\mathbf{v}}}\mathbf{=}\frac{{\displaystyle \frac{\mathbf{5}\mathbf{R}}{\mathbf{2}}}}{{\displaystyle \frac{\mathbf{3}\mathbf{R}}{\mathbf{2}}}} \end{gathered}$$

Now, the ratio for given gas,

$\frac{{\mathbf{C}}_{\mathbf{P}}}{{\mathbf{C}}_{\mathbf{v}}}\mathbf{=}\frac{\mathbf{23}\mathbf{.}\mathbf{1}}{\mathbf{20}\mathbf{.}\mathbf{79}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{11}$

Since, the two values does not match. Thus the given gas is not monatomic.