Question

Two identical samples of gas are expanded so that the volume is increased to twice the initial volume. However, sample number 1 is expanded isothermally while sample number 2 is expanded adiabatically. In which sample is the pressure greater? Why?​

Answer 1

Answer:

nothing can be said. Since slope of adiabatic graph is γ-times more than slope of isothermal process. These final pressure is more is isothermal process than adiabatic process. Thus, correct choice is (a).

Answer 2

Answer:

Sample number 1 will have greater pressure as compared to sample 2.

Explanation:

Isothermal process:

• It is a thermodynamic process in which the temperature is kept constant.
• An isothermal expansion is a process where the gas is expanded in volume keeping the temperature constant.
• If we consider the gas to be ideal, then the ideal gas equation is given by,

                               $PV=nRT$

        Where 'P' is the pressure of the gas, 'V' is the volume of the gas, 'n'          is the number of moles of the gas, 'R' is the universal gas constant and 'T' is the temperature of the gas.

Since 'T' is constant in an isothermal process the RHS of the ideal gas equation is a constant for a fixed amount of gas. From this, we can conclude

                            $PV=constant$

• If the gas is expanded from a volume $V_{1}$ to $V_{2}$ and the pressure changes from $P_{1}$ to $P_{2}$ then from the above equation we have,

                           $P_{1} V_{1} = P_{2} V_{2}$

• We can also understand that in an isothermal process the pressure is inversely proportional to the volume.

Adiabatic Process:

• It is a thermodynamic process in which there is no transfer of heat and mass between the system and the surrounding.
• The mathematical equation for an ideal gas undergoing reversible adiabatic process is given by

                           $PV^{\gamma} = constant$

Where $\gamma$ is the ratio of specific heat at constant pressure to specific heat at constant volume.

• $\gamma$ is given by

                          $\gamma = \frac{C_{P} }{C_{V} }$

        Where $C_{P}$ is the specific heat at constant pressure and $C_{V}$ is the specific heat at constant volume.

        For a gas $C_{P} >C_{V}$, therefore $\gamma>1$. For monoatomic gas $\gamma = 1.66$ and for diatomic gas $\gamma=1.4$.

Step 1:

Given in the question two identical samples of gas are expanded in two different ways. One is expanded isothermally to twice its volume and the other expanded adiabatically to twice its volume.

For sample number 1:

Initial pressure = $P_{1}$

Final pressure = $P_{2}$

Initial volume = $V_{1}$

Final volume =$2V_{1}$

Similarly, for sample number 2:

Initial pressure = $P_{1}$

Final pressure = $P_{2}$

Initial volume = $V_{1}$

Final volume =$2V_{1}$

Step 2:

First, let's find the final pressure of sample 1 after isothermal expansion to twice its volume. As mentioned earlier the equation for an ideal gas undergoing isothermal expansion is given by,

               $PV=const\\P_{1} V_{1} = P_{2} V_{2} \\P_{1} V_{1} = P_{2} (2V_{1})\\ P_{2} = \frac{P_{1}}{2}$

Here, the final pressure after expansion becomes half of the initial pressure.

Step 3:

Now, we will calculate the final pressure of sample 2 after adiabatic expansion to twice its volume. As mentioned earlier the equation for an ideal gas undergoing adiabatic expansion is given by,

               $PV^{\gamma} = constant\\P_{1} V_{1} ^{\gamma} = P_{2} V_{2} ^{\gamma}\\ P_{1} V_{1} ^{\gamma} = P_{2} (2V_{1}) ^{\gamma}\\P_{2} = \frac{P_{1}}{2^{\gamma} }$

As mentioned earlier $\gamma$ for a gas is always greater than 1, hence in the equation for the final pressure, there is an $2^{\gamma}$ in the denominator which is always greater than 2. That is

               $2^{\gamma} > 2$

Due to the presence of $2^{\gamma}$ in the denominator, the pressure after expansion in an isothermal process is greater than that in an adiabatic process.