Q.Consider the expansion of 1 mol of an ideal gas (from 2 L to 10 L at 25 °C) conducted reversibly.

The problem asks us to find the temperature $T_2$ given $T_1$ and the ratio of slopes of isotherms plotted as $p$ versus $1/V$. To solve this, we need to understand how the ideal gas law relates to such a graph.

The ideal gas law describes the relationship between pressure ($P$), volume ($V$), temperature ($T$), and the number of moles ($n$) of an ideal gas:

An isotherm is a curve representing the states of a gas at a constant temperature. When we plot $p$ against $1/V$, we are essentially looking for a relationship of the form $y = mx + c$, where $y=p$ and $x=1/V$.

Let's rearrange the ideal gas law to match this form:

$P = \frac{nRT}{V}$

$P = (nRT) \left(\frac{1}{V}\right)$

Comparing this to the equation of a straight line, $y = mx + c$:

• $y$ corresponds to $P$ (pressure)
• $x$ corresponds to $\frac{1}{V}$ (inverse of volume)
• The intercept $c$ is $0$, meaning the line passes through the origin.
• The slope $m$ corresponds to $nRT$.

Since $n$ (number of moles) and $R$ (ideal gas constant) are constant for a given amount of gas, the slope of the $p$ versus $1/V$ isotherm is directly proportional to the absolute temperature $T$.

Now, let's apply this understanding to the given problem.

1. Express the slope in terms of temperature:

   For an ideal gas, the equation relating pressure $p$ and inverse volume $1/V$ at a constant temperature $T$ is derived from the ideal gas law:

   $p = (nRT) \frac{1}{V}$

   This is in the form $y = mx$, where $y=p$ and $x=1/V$.

   The slope of this line is $m = nRT$.

2. Write the slopes for $T_1$ and $T_2$:

   Let $m_1$ be the slope of the isotherm at temperature $T_1$, and $m_2$ be the slope of the isotherm at temperature $T_2$.

   $m_1 = nRT_1$

   $m_2 = nRT_2$ …

The concept first: why work is an area on a $P$–$V$ plot

When a gas in a cylinder is compressed against an external pressure $P_{\text{ext}}$, pushing the piston in by a tiny amount changes the volume by $\mathrm{d}V$ and the work done on the gas is

The minus sign is the sign convention: compression means $\mathrm{d}V<0$, so $\mathrm{d}w>0$ — work is done on the gas, its energy rises.

For a reversible change the external pressure is only infinitesimally different from the gas pressure, so $P_{\text{ext}} \simeq P$ (the pressure plotted on the graph), and summing all the little slices:

A definite integral of $P$ with respect to $V$ is, geometrically, the area lying between the curve $P(V)$ and the $V$-axis, cut off by the vertical lines $V=V_1$ and $V=V_2$. That is the whole reason a $P$–$V$ diagram is so useful: work is an area on it.

For the isothermal ideal gas drawn in the stem, $P = nRT/V$, and the integral evaluates to the familiar

Step-by-step: test each shaded region

1. What must the region look like? Four boundaries: the isotherm on top, the $V$-axis below, and two vertical lines at the initial volume $V_1$ and the final volume $V_2$. No process runs from $V=0$, and no process has an undefined end point. …

Initial volume of the gas:

Compressed to half its volume:

Work done on the gas against the constant external pressure $P_{ext} = 6.0$ atm: …

The key concept is the formula for reversible isothermal work on an ideal gas. When a gas expands reversibly at constant temperature, the pressure changes continuously, so we must integrate $P\,dV$. Using the ideal gas law $PV = nRT$, we replace $P$ with $nRT/V$, leading to $W = -nRT \ln(V_f/V_i)$. Since pressure and volume are inversely related at constant temperature ($P_i V_i = P_f V_f$), we can also write $W = -nRT \ln(P_i/P_f)$. This form is convenient here because we are given pressures directly.

Now, step by step:

1. Identify the formula. For a reversible isothermal expansion of an ideal gas:

The negative sign indicates work is done by the system (expansion).

2. Plug in the values.

   • $n = 2$ mol
   • $R = 8.3$ J K$^{-1}$ mol$^{-1}$
   • $T$ is the temperature in Kelvin (kept as a variable)
   • $P_i = 10$ atm, $P_f = 1$ atm, so $\frac{P_i}{P_f} = 10$.

   Thus:

3. Compute $\ln(10)$. $\ln(10) \approx 2.302585$. Multiply:

So $W \approx -38.22 \times T$ joules.

4. Convert to kilojoules. …

Concept and Intuition

The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) for a process is given by the first law of thermodynamics applied to constant pressure conditions. Enthalpy is defined as $H = U + PV$. Therefore, for a process occurring at constant pressure $P$, the change in enthalpy is:

Here, $\Delta U$ is the change in internal energy, $P$ is the constant pressure, and $\Delta V$ is the change in volume of the system. The term $P\Delta V$ represents the pressure-volume work done by or on the system.

• For reactions involving gases: The volume changes ($\Delta V$) can be substantial, and thus the $P\Delta V$ term is often significant and cannot be ignored. For ideal gases, $P\Delta V = \Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gas.
• For reactions involving only liquids and solids (condensed phases): The volume changes are typically very small. Consequently, the $P\Delta V$ term is usually negligible compared to $\Delta U$. In such cases, it is a common and often accurate approximation to state that $\Delta H \approx \Delta U$.

In this problem, we are dealing with the phase transition of water from liquid to solid (freezing). While water is unique in that its volume increases upon freezing (ice is less dense than liquid water), this volume change is still very small compared to the volume changes observed in gas-phase reactions. We need to calculate the $P\Delta V$ term to see if it's significant.

Step-by-Step Solution

1. Identify the given information and the reaction:

   The reaction is the freezing of water:

   $\mathrm{H_2O(l) \rightarrow H_2O(s)}$

   The conditions are $0\,^\circ\mathrm{C}$ and $1$ atm.

   The internal energy change is $\Delta U = -41$ kJ/mol.

   We need to find the molar enthalpy change, $\Delta H$.

2. Recall the fundamental relationship:

   As discussed, the relationship between $\Delta H$ and $\Delta U$ at constant pressure is:

   $\Delta H = \Delta U + P\Delta V$

3. Determine the change in molar volume ($\Delta V$):

   To calculate $P\Delta V$, we need the molar volumes of liquid water and ice at $0\,^\circ\mathrm{C}$ and $1$ atm. We use standard density values:

   • Molar mass of water ($M$) = $18.015$ g/mol
   • Density of liquid water ($\rho_l$) at $0\,^\circ\mathrm{C}$ = $0.9998$ g/cm$^3$
   • Density of ice ($\rho_s$) at $0\,^\circ\mathrm{C}$ = $0.9167$ g/cm$^3$

   Calculate the molar volume of liquid water ($V_l$):

   $V_l = \frac{M}{\rho_l} = \frac{18.015 \text{ g/mol}}{0.9998 \text{ g/cm}^3} = 18.0186 \text{ cm}^3\text{/mol}$

   Calculate the molar volume of ice ($V_s$):

   $V_s = \frac{M}{\rho_s} = \frac{18.015 \text{ g/mol}}{0.9167 \text{ g/cm}^3} = 19.651 \text{ cm}^3\text{/mol}$

   Now, calculate the change in molar volume ($\Delta V$):

   $\Delta V = V_s - V_l = 19.651 \text{ cm}^3\text{/mol} - 18.0186 \text{ cm}^3\text{/mol} = 1.6324 \text{ cm}^3\text{/mol}$ …

Step 1 — reaction and moles of gas.

Step 2 — work of expansion. In an open vessel the gas expands against constant atmospheric pressure, so

$w=-P\,\Delta V=-\Delta n_g\,RT,$ …