Question

A mixture of an ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at a pressure ${{P}_{0}}$, volume ${{V}_{0}}$, and temperature ${{T}_{0}}$. If the mixture is adiabatically compressed to a volume ${{V}_{0}}/4$, then the correct statements are:
Given information: ${{2}^{1.2}}=2.3$, ${{2}^{3.2}}=9.2$, and R is the gas constant.
A. The final pressure of the gas mixture after compression is between $9{{P}_{0}}$ and $10{{P}_{0}}$
B. The average kinetic energy of the gas mixture after compression is in between $18R{{T}_{0}}$ and $19R{{T}_{0}}$
C. Adiabatic constant of the gas mixture is 1.6
D. The work done $\left| w \right|$ during the process is $13R{{T}_{0}}$

Answer 1

Think about the formulae that are used to calculate the parameters that are given in each of the options. Calculate the degrees of freedom from the given information on whether the gas is monatomic or diatomic.

Complete step by step solution:
We will look at each of the given options one by one, calculate the values and check whether the given statements are true or not.
- Option A
For this, we need to calculate the pressure exerted by the gas after the compression has occurred. To do this, we will use the formula that involves the pressure, volume and the adiabatic constant $\gamma$ of the gas mixture. We will use the formula:
$i){{P}_{1}}{{V}_{1}}^{\gamma }={{P}_{2}}{{V}_{2}}^{\gamma }$
Where, ${{P}_{1}}$ and ${{V}_{1}}$ are the initial pressure and volumes of the combination of the gases, ${{P}_{2}}$ and ${{V}_{2}}$ are the final pressure and volumes of the combination of the gases, and $\gamma$ is the adiabatic constant of the mixture of gases. We know that the formula for the adiabatic constant is:
$ii)\gamma =1+\dfrac{2}{{{f}_{mix}}}$
Here, ${{f}_{mix}}$ shows the mixed degree of freedom of both the gases when they are combined, to find this, we use the formula given below.
$iii){{f}_{mix}}=\dfrac{{{n}_{1}}{{f}_{1}}+{{n}_{2}}{{f}_{2}}}{{{n}_{1}}+{{n}_{2}}}$
Here, $n$ denotes the number of moles of gases 1 and 2 and $f$ denotes the degree of freedom of gases 1 and 2. So, from the information given in the question, we know the following data:

& {{n}_{1}}=5mol \\\ & {{n}_{2}}=1mol \\\ & {{f}_{1}}=3 \\\ & {{f}_{2}}=5 \\\ \end{aligned}$$ Putting these values in the formula ii) to find the mixed degree of freedom, we get: $$\begin{aligned} & \Rightarrow {{f}_{mix}}=\dfrac{5\times 3+1\times 5}{5+1} \\\ & \Rightarrow {{f}_{mix}}=\dfrac{15+5}{6} \\\ & \Rightarrow {{f}_{mix}}=\dfrac{10}{3} \\\ \end{aligned}$$ Now, putting this value of ${{f}_{mix}}$ in the formula to find the adiabatic constant, we get: $$\begin{aligned} & \gamma =1+\dfrac{2}{\dfrac{10}{3}} \\\ &\Rightarrow \gamma =1+\dfrac{6}{10}=\dfrac{16}{10} \\\ &\Rightarrow \gamma =1.6 \\\ \end{aligned}$$ Now that we have this value, we can use it to find the final pressure exerted by the mixture of gases. The values that are known to us from the question are: $$\begin{aligned} & {{P}_{1}}={{P}_{0}} \\\ & {{V}_{1}}={{V}_{0}} \\\ & {{V}_{2}}=\dfrac{{{V}_{0}}}{4} \\\ \end{aligned}$$ Putting these values along with the value of $\gamma $ in formula i) to find the final pressure, we get: $$\begin{aligned} & {{P}_{0}}{{V}_{0}}^{1.6}={{P}_{2}}{{\left( \dfrac{{{V}_{0}}}{4} \right)}^{1.6}} \\\ &\Rightarrow {{P}_{2}}={{P}_{0}}\times {{\left( \dfrac{4\times {{V}_{0}}}{{{V}_{0}}} \right)}^{1.6}} \\\ & \Rightarrow {{P}_{2}}={{P}_{0}}\times {{4}^{1.6}} \\\ & \Rightarrow {{P}_{2}}={{2}^{3.2}}{{P}_{0}} \\\ & \Rightarrow {{P}_{2}}=9.2{{P}_{0}} \\\ \end{aligned}$$ Here, we can see that the final pressure is 9.2 times the initial pressure and that the final pressure after compression does lie in between $9{{P}_{0}}$ and $10{{P}_{0}}$. Thus, option A is true. \- Option B The formula for the average kinetic energy after the adiabatic compression is defined as: $$iv)KE=\dfrac{{{f}_{mix}}}{2}nR{{T}_{2}}$$ Here, $n$ is the total number of moles of gas that are present and ${{T}_{2}}$ is the final temperature after the compression. We need to find the value of the final temperature, to do that, we will use formula i) and carry out substitutions using the ideal gas equation. So the formula we will use to find the final temperature will be: $$v){{T}_{1}}{{V}_{1}}^{\gamma -1}={{T}_{2}}{{V}_{2}}^{\gamma -1}$$ We will now substitute the values that we have in the above equation and obtain the value of ${{T}_{2}}$. The value of ${{T}_{1}}$ is ${{T}_{0}}$. Solving for ${{T}_{2}}$ we get: $$\begin{aligned} & {{T}_{0}}{{V}_{0}}^{1.6-1}={{T}_{2}}{{\left( \dfrac{{{V}_{0}}}{4} \right)}^{1.6-1}} \\\ & \Rightarrow {{T}_{2}}={{T}_{0}}\times {{\left( \dfrac{{{V}_{0}}\times 4}{{{V}_{0}}} \right)}^{0.6}} \\\ & \Rightarrow {{T}_{2}}={{T}_{0}}\times {{4}^{0.6}} \\\ & \Rightarrow {{T}_{2}}={{2}^{1.2}}{{T}_{0}} \\\ & \Rightarrow {{T}_{2}}=2.3{{T}_{0}} \\\ \end{aligned}$$ Now, we will put the obtained values in the formula iv) and then solve the equation for the average kinetic energy. The equation will be as follows: $$\begin{aligned} & KE=\dfrac{10}{6}\times 6\times R\times 2.3{{T}_{0}} \\\ & \Rightarrow KE=23R{{T}_{0}} \\\ \end{aligned}$$ As given in option B, this value does not lie in between $18R{{T}_{0}}$ and $19R{{T}_{0}}$. Thus, option B is not true. \- Option C The adiabatic constant of the gas mixture is denoted by the variable $\gamma $ which we have already obtained from formula ii) as 1.6. Thus, option C is true. \- Option D The general formula for work done in an adiabatically expanding gas is given by: $$vi)\left| w \right|=\dfrac{nR\left( {{T}_{2}}-{{T}_{1}} \right)}{\gamma -1}$$ We already have the values that are required for this calculation form the previous calculations and all the variables take their previously defined values. Solving for $\left| w \right|$, we get: $$\begin{aligned} & \left| w \right|=\dfrac{6\times R\left( 2.3{{T}_{0}}-{{T}_{0}} \right)}{1.6-1} \\\ & \Rightarrow \left| w \right|=\dfrac{6\times 1.3R{{T}_{0}}}{0.6} \\\ & \Rightarrow \left| w \right|=10\times 1.3{{T}_{0}} \\\ & \Rightarrow \left| w \right|=13R{{T}_{0}} \\\ \end{aligned}$$ So, we have the value of the work done as $13R{{T}_{0}}$ and thus, the statement in option D is true. **Hence, the correct options in this question are option A, option C, and option D.** **Note:** Remember that for the work done, we are taking the modulus of the value since for adiabatic compression, the work done by the gas will always be negative. Since, we do not want a negative sign, we will consider the modulus of the work done.