Q.An ideal gas is carried once around the cyclic process $A\to B\to C\to D\to A$ on a $P$-$V$ diagram whose four corners are: $A$ at (volume $V_o$, pressure $P_o$), $B$ at (volume $3V_o$, pressure $P_o$), $C$ at (volume $3V_o$, pressure $2P_o$) and $D$ at (volume $V_o$, pressure $2P_o$). The steps are: $A\to B$ an expansion at constant pressure $P_o$; $B\to C$ a pressure rise at constant volume $3V_o$; $C\to D$ a compression at constant pressure $2P_o$; $D\to A$ a pressure drop at constant volume $V_o$. Find the net work done by the gas in one complete cycle.

The problem asks for the work done during the adiabatic expansion phase of a Carnot engine. In an adiabatic process, there is no heat exchange with the surroundings. The work done by the gas in such a process comes entirely from the change in its internal energy.

For an ideal gas, the change in internal energy ($\Delta U$) is given by $nC_v\Delta T$, where $n$ is the number of moles, $C_v$ is the molar specific heat capacity at constant volume, and $\Delta T$ is the change in temperature. According to the first law of thermodynamics, $dQ = dU + dW$. For an adiabatic process, $dQ = 0$, so $dW = -dU$. This means the work done by the gas ($dW$) is equal to the negative of the change in internal energy. If the gas expands, it does positive work, and its internal energy decreases (temperature drops).

The key to solving this problem is to first determine the specific heat capacity $C_v$ for the working substance, as it's not a standard monatomic or diatomic gas. We are given a specific adiabatic relation $T\sqrt{V} = \text{constant}$, which we can use to find the adiabatic index $\gamma$. Once $C_v$ is known, we can calculate the work done using the temperature difference between the hot and cold reservoirs, as adiabatic expansion in a Carnot cycle occurs between these two temperatures.

1. Convert temperatures to absolute scale:

   Thermodynamic calculations involving temperature differences or ratios must use absolute temperatures (Kelvin).

   The higher temperature (hot reservoir) is $T_H = 133\,^\circ\mathrm{C} = 133 + 273 = 406\,\mathrm{K}$.

   The lower temperature (cold reservoir) is $T_C = 33\,^\circ\mathrm{C} = 33 + 273 = 306\,\mathrm{K}$.

2. Determine the adiabatic index ($\gamma$) for the working substance:

   The general relation between absolute temperature ($T$) and volume ($V$) for an adiabatic process involving an ideal gas is:

We are given the specific relation for this working substance:

This can be written as:

Comparing this with the general adiabatic relation, we can see that:

Therefore, the adiabatic index for this working substance is:

3. Calculate the molar specific heat capacity at constant volume ($C_v$): For an ideal gas, the molar specific heat capacities at constant pressure ($C_p$) and constant volume ($C_v$) are related by: …

Concept and Intuition

This problem involves the thermodynamics of an ideal gas undergoing an isobaric process, meaning the pressure remains constant. To find the heat supplied, we use the First Law of Thermodynamics, which states that the heat supplied to a system ($Q$) is used to increase its internal energy ($\Delta U$) and to do work ($W$) on the surroundings.

Let's break down each term for an isobaric process:

1. Work Done ($W$): When a gas expands against a constant external pressure $P$, the work done by the gas is given by $W = P\Delta V$, where $\Delta V$ is the change in volume.
2. Change in Internal Energy ($\Delta U$): For an ideal gas, the internal energy depends only on its temperature. The change in internal energy is given by $\Delta U = nC_v\Delta T$, where $n$ is the number of moles, $C_v$ is the molar specific heat at constant volume, and $\Delta T$ is the change in temperature.
3. Heat Supplied ($Q$): Substituting these into the First Law:

For an ideal gas, the ideal gas law states $PV = nRT$. If the pressure $P$ is constant, then $P\Delta V = nR\Delta T$.
Substituting $P\Delta V = nR\Delta T$ into the equation for $Q$:

We know that for an ideal gas, the molar specific heat at constant pressure, $C_p$, is related to $C_v$ and the universal gas constant $R$ by Mayer's relation: $C_p = C_v + R$.
Therefore, for an isobaric process, the heat supplied can be directly calculated as:

> [!FORMULA]
> For an ideal gas undergoing an isobaric process, the heat supplied $Q$ is given by:
> $$Q = nC_p\Delta T$$

Now, we need to determine $C_p$ for a diatomic gas.
An ideal diatomic gas has 5 degrees of freedom at moderate temperatures (3 translational and 2 rotational).
The molar specific heat at constant volume is $C_v = \frac{f}{2}R$, where $f$ is the degrees of freedom.
For a diatomic gas, $f=5$, so $C_v = \frac{5}{2}R$.
Using Mayer's relation, $C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$.

So, for an ideal diatomic gas undergoing an isobaric process, the heat supplied is:

Again, using the ideal gas law for an isobaric process, $P\Delta V = nR\Delta T$.
Substituting this into the expression for $Q$:

This formula allows us to calculate the heat supplied directly from the given pressure and change in volume, without needing to know the number of moles or the temperature change explicitly.

Step-by-step Solution

1. Identify the given values and convert units:

   • Initial volume, $V_1 = 18 \times 10^{-3} \ \mathrm{m}^3$.
   • Final volume, $V_2 = 37 \times 10^{-3} \ \mathrm{m}^3$.
   • Constant pressure, $P = 2 \ \mathrm{atm}$.
   • Atmospheric pressure conversion: $1 \ \mathrm{atm} = 10^5 \ \mathrm{Pa}$.
   • Therefore, $P = 2 \times 10^5 \ \mathrm{Pa}$.

2. Calculate the change in volume ($\Delta V$):

   The change in volume is $\Delta V = V_2 - V_1$. …

The concept: In thermodynamics, we distinguish between state variables (like pressure, volume, temperature, internal energy) which depend only on the current state of the system, and path functions (like heat and work) which depend on how you get from one state to another. Heat and work are not properties of the system itself — they are energy in transit. The first law of thermodynamics, $\Delta U = Q + W$, tells us that the sum of heat and work equals the change in internal energy, but individually each depends on the path taken.

1. Assertion (A) is true. Heat and work are the two ways energy can cross the boundary of a thermodynamic system. When you add heat to a gas or do work on it by compressing it, the internal energy changes. This is exactly what the first law states: the change in internal energy equals the net energy transferred as heat and work. …

Concept & intuition

Heat is path-dependent, so use $Q=\Delta U+W$. Internal energy depends only on the endpoints, while the work is the area under the linear $P$–$V$ path.

Step-by-step solution

1. Change in internal energy. For a monoatomic ideal gas $U=\tfrac32 PV$, so

2. Work done by the gas (area of the trapezoid under the straight line from $(V_o,P_o)$ to $(3V_o,3P_o)$): $W=\underbrace{\tfrac{P_o+3P_o}{2}}_{\text{avg pressure}=2P_o}\times\underbrace{(3V_o-V_o)}_{2V_o}=2P_o\cdot 2V_o=4P_oV_o.$ …

When a liquid drop is broken into smaller drops, the total surface area of the liquid increases. Since liquid surfaces possess surface energy due to surface tension, an increase in surface area means an increase in surface energy. This additional energy must come from an external source, which is the work done in breaking the drop.

The fundamental principle here is that work done ($W$) in changing the surface area of a liquid is given by the product of the surface tension ($T$) and the change in surface area ($\Delta A$).

$W = T \times \Delta A$

To calculate $\Delta A$, we need to determine the initial surface area of the large drop and the total final surface area of all the smaller drops. A crucial point is that the total volume of the liquid remains constant throughout the process.

1. Relate the radii using volume conservation:

   Let the radius of the large drop be $R$ and its volume be $V_1$.

   $V_1 = \frac{4}{3}\pi R^3$

   When this drop breaks into 64 equal smaller drops, let the radius of each small drop be $r$. The volume of one small drop is $v = \frac{4}{3}\pi r^3$.

   The total volume of the 64 small drops is $V_2 = 64 \times \frac{4}{3}\pi r^3$.

   By the principle of conservation of volume:

   $V_1 = V_2$

   $\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3$

   $R^3 = 64 r^3$

   Taking the cube root of both sides:

   $R = \sqrt[3]{64} r$

   $R = 4r$

   This gives us the relationship between the radii: $r = \frac{R}{4}$.

2. Calculate the initial surface area:

   The surface area of the original large drop, $A_1$, is given by the formula for the surface area of a sphere:

   $A_1 = 4\pi R^2$

3. Calculate the final total surface area:

   The surface area of one small drop, $a$, is $4\pi r^2$.

   Substituting $r = \frac{R}{4}$:

   $a = 4\pi \left(\frac{R}{4}\right)^2 = 4\pi \frac{R^2}{16} = \frac{\pi R^2}{4}$

   Since there are 64 such small drops, the total final surface area, $A_2$, is: …