The work of 146 KJ in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by ${7^o}C$. The gas is $\left( {R = 8.3Jmo{l^{ - 1}}{K^{ - 1}}} \right)$
A) Monoatomic
B) Diatomic
C) Triatomic
D) A mixture of monoatomic and diatomic
Answer
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Hint: To solve this problem one must be aware of the concept of adiabatic index. The adiabatic index of a gas is the ratio of the specific heat of constant pressure of the gas to its specific heat at constant volume. In order to find if the gas is monatomic, diatomic or triatomic we must find out its adiabatic index $(\gamma )$.
Complete step by step answer:
It is given in the problem that the amount of gas is 1 kilo mole, i.e. $n = 1 \times {10^3}mol$
Also, the work done on the gas is 146 KJ, i.e. $W = 146 \times {10^3}J$
The rise in temperate is given to be ${7^o}C$, i.e. $\Delta T = {7^o}C$
We know that work done for a closed system in an adiabatic process is given by,
$W = \dfrac{{nR\Delta T}}{{\gamma - 1}}$
Hence, from the above equation we can derive an expression to find the value of adiabatic index and it is given below,
$ \Rightarrow \gamma = 1 + \dfrac{{nR\Delta T}}{W}$ ………. (1)
Putting all the given values in equation (1), we get,
$\gamma = 1 + \dfrac{{{{10}^3} \times 8.3 \times 7}}{{146 \times {{10}^3}}}$
$ \Rightarrow \gamma = 1.39$
$\therefore \gamma \approx 1.4$
Here we have found that the value of adiabatic index is approximately equal to 1.4. We know that a gas with an adiabatic index of value 1.4 is said to be a diatomic gas.
Hence, option B is the correct answer option.
Note: Following key points regarding the adiabatic index must be kept in mind.
Adiabatic index has no unit as it is a ratio of two similar quantities.
1. If adiabatic index of a gas has a value of 1.66, i.e $\gamma = 1.66$ then the gas is said to be monatomic.
2. If the adiabatic index of a gas has a value of 1.4, i.e $\gamma = 1.4$ then the gas is said to be diatomic.
3. If the adiabatic index of a gas has a value of 1.33, i.e $\gamma = 1.33$ then the gas is said to be triatomic.
Complete step by step answer:
It is given in the problem that the amount of gas is 1 kilo mole, i.e. $n = 1 \times {10^3}mol$
Also, the work done on the gas is 146 KJ, i.e. $W = 146 \times {10^3}J$
The rise in temperate is given to be ${7^o}C$, i.e. $\Delta T = {7^o}C$
We know that work done for a closed system in an adiabatic process is given by,
$W = \dfrac{{nR\Delta T}}{{\gamma - 1}}$
Hence, from the above equation we can derive an expression to find the value of adiabatic index and it is given below,
$ \Rightarrow \gamma = 1 + \dfrac{{nR\Delta T}}{W}$ ………. (1)
Putting all the given values in equation (1), we get,
$\gamma = 1 + \dfrac{{{{10}^3} \times 8.3 \times 7}}{{146 \times {{10}^3}}}$
$ \Rightarrow \gamma = 1.39$
$\therefore \gamma \approx 1.4$
Here we have found that the value of adiabatic index is approximately equal to 1.4. We know that a gas with an adiabatic index of value 1.4 is said to be a diatomic gas.
Hence, option B is the correct answer option.
Note: Following key points regarding the adiabatic index must be kept in mind.
Adiabatic index has no unit as it is a ratio of two similar quantities.
1. If adiabatic index of a gas has a value of 1.66, i.e $\gamma = 1.66$ then the gas is said to be monatomic.
2. If the adiabatic index of a gas has a value of 1.4, i.e $\gamma = 1.4$ then the gas is said to be diatomic.
3. If the adiabatic index of a gas has a value of 1.33, i.e $\gamma = 1.33$ then the gas is said to be triatomic.
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