Answer
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Hint: An ideal gas when heated at constant volume, the work done is zero. The entire het will be used in raising the kinetic energy. Also for isochoric processes the pressure will change in order to keep the volume constant.
Complete step by step solution:
In the given condition the volume is constant. The change in volume is zero. The pressure is changing to keep the volume constant. The entire heat is converted to the internal energy as no work is done.
So let us apply the first law of thermodynamics;
\[\Delta Q = \Delta U + W\]
We also know that the work done in a thermodynamic process is given by;
$W = p\Delta V$ Here P is the pressure and $\Delta V$ is the change in volume.
Thus, $W = 0$
Hence it is clear that the work done in an isochoric process is zero.
So the first law for isochoric process is given by;
$\Delta Q = \Delta U$
And so, the entire energy supplied as heat is used to raise the internal energy of the system.
Now, according to the gas law;
$PV = nRT$
$ \Rightarrow P = \dfrac{{nRT}}{V}$ (Equation: 1)
Thus if we increase the temperature the pressure increases.
This tells us that pressure is directly proportional to temperature. Thus mathematically;
$P$ $\alpha $ $T$
Thus if temperature increases the pressure exerted on the wall also increases.
Now according to kinetic theory of gases, the kinetic energy of gas is given by;
$KE = \dfrac{3}{2}kT$
Here k is the bulk modulus, and T is the temperature.
The above equation also shows that the kinetic energy of the gas is directly proportional to the temperature.
Thus we conclude that the kinetic energy of the gas increases with the rise in temperature.
Hence statement I and statement II are correct.
Therefore option B is correct.
Note:The average kinetic energy of the gas would have increased even if the volume was not constant, however the value may be less.
This is because the average kinetic energy does not depend on volume and pressure. So, it is unaffected by the process being isobaric or isochoric.
The average kinetic energy is dependent on temperature and hence is constant for isothermal processes.
Complete step by step solution:
In the given condition the volume is constant. The change in volume is zero. The pressure is changing to keep the volume constant. The entire heat is converted to the internal energy as no work is done.
So let us apply the first law of thermodynamics;
\[\Delta Q = \Delta U + W\]
We also know that the work done in a thermodynamic process is given by;
$W = p\Delta V$ Here P is the pressure and $\Delta V$ is the change in volume.
Thus, $W = 0$
Hence it is clear that the work done in an isochoric process is zero.
So the first law for isochoric process is given by;
$\Delta Q = \Delta U$
And so, the entire energy supplied as heat is used to raise the internal energy of the system.
Now, according to the gas law;
$PV = nRT$
$ \Rightarrow P = \dfrac{{nRT}}{V}$ (Equation: 1)
Thus if we increase the temperature the pressure increases.
This tells us that pressure is directly proportional to temperature. Thus mathematically;
$P$ $\alpha $ $T$
Thus if temperature increases the pressure exerted on the wall also increases.
Now according to kinetic theory of gases, the kinetic energy of gas is given by;
$KE = \dfrac{3}{2}kT$
Here k is the bulk modulus, and T is the temperature.
The above equation also shows that the kinetic energy of the gas is directly proportional to the temperature.
Thus we conclude that the kinetic energy of the gas increases with the rise in temperature.
Hence statement I and statement II are correct.
Therefore option B is correct.
Note:The average kinetic energy of the gas would have increased even if the volume was not constant, however the value may be less.
This is because the average kinetic energy does not depend on volume and pressure. So, it is unaffected by the process being isobaric or isochoric.
The average kinetic energy is dependent on temperature and hence is constant for isothermal processes.
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