Question

What is the increment of the internal energy of the gas?
A.\dfrac{{15\alpha {V_0}^2}}{{\gamma - 1}} \\\ B.\dfrac{{\alpha {V_0}^2}}{{\gamma - 1}} \\\ C.\dfrac{{15\alpha {V_0}^2}}{2} \\\ D.\dfrac{{15\alpha (\gamma + 1){V_0}^2}}{{2(\gamma - 1)}} \\\

Answer 1

Hint : Change in internal energy of a gas is the sum of heat absorbed or given out of the gas since the process is adiabatic and the work done by the gas. Work done is in changing the volume of the system. While compressing the gas two things are done, heat is absorbed and temperature changes.
$PV = nRT$ That is the basic equation used in the kinetic of gases.
Now,
{V_0} = \dfrac{{{V_f}}}{4} \\\ \dfrac{{{P_0}}}{{{V_0}}} = const \\\

Complete Step By Step Answer:
In order to solve this question we first see that,
For adiabatic process of expansion;
Taking alpha as constant,
Pressure is equal to alpha times the volume of the gas.
So, ${P_0} = \alpha {V_0}$
And, ${V_f} = 4{V_0}$
Therefore,
\vartriangle U = n{C_v}dT \\\ = n \times \dfrac{R}{{(\gamma - 1)}}\vartriangle T \\\ = \dfrac{{nR({T_f} - {T_0})}}{{(\gamma - 1)}} \\\
Since,
$PV = nRT$
Now,
{V_0} = \dfrac{{{V_f}}}{4} \\\ \dfrac{{{P_0}}}{{{V_0}}} = const \\\
Therefore;
$\dfrac{{{P_0}}}{{{V_0}}} = \dfrac{{{P_f}}}{{{V_f}}} = const$
Now combining all these we get;
$\dfrac{{{P_0}}}{{{P_f}}} = \dfrac{{{V_0}}}{{{V_f}}} = \dfrac{1}{4}$
Therefore,
{P_f} = 4\alpha {V_0} \\\ {V_f} = 4{V_0} \\\
Now substituting all the values hereby;
\vartriangle U = \dfrac{{4\alpha {V_0}(4{V_0}) - \alpha {V_0}({V_0})}}{{(\gamma - 1)}} \\\ = \dfrac{{16\alpha {V_0}^2 - \alpha {V_0}^2}}{{(\gamma - 1)}} \\\ = \dfrac{{15\alpha {V_0}^2}}{{(\gamma - 1)}} \\\
Hence option A is the correct answer.

Note :
Internal energy is the sum of all the kinetic energy performed by the gas around the molecule. Change in internal energy of a gas is the sum of heat absorbed or given out of the gas since the process is adiabatic and the work done by the gas. Work done is in changing the volume of the system. While compressing the gas two things are done, heat is absorbed and temperature changes. Since it’s the ideal gas, all the energy is in the form of kinetic energy.
Internal energy of a gas may be increased when changed in adiabatic.
If we are talking about the ratio, then;
{V_0} = \dfrac{{{V_f}}}{4} \\\ \dfrac{{{P_0}}}{{{V_0}}} = const \\\
Therefore;
$\dfrac{{{P_0}}}{{{V_0}}} = \dfrac{{{P_f}}}{{{V_f}}} = const$