Concept & Intuition

When equal amounts of heat are supplied at constant pressure, the temperature rise depends on the molar specific heat at constant pressure, $C_p$. For an ideal gas, the change in internal energy $\Delta U$ depends only on the temperature change and the molar specific heat at constant volume, $C_v$. The ratio of $\Delta U$ for two gases thus involves both $C_p$ (to relate heat to $\Delta T$) and $C_v$ (to relate $\Delta T$ to $\Delta U$). The key is that monoatomic and diatomic gases have different degrees of freedom, giving different $C_v$ and $C_p$ values.

Step-by-step solution

1. Recall the relevant specific heats For an ideal monoatomic gas:

For a rigid diatomic gas (no vibration, only translation and rotation):

2. Relate heat supplied to temperature change at constant pressure At constant pressure, the heat $Q$ supplied to $n$ moles is:

Since equal amounts of heat are supplied to equal numbers of moles (implied by "equal amounts of heat" and same amount of gas), we have:

Cancelling $n$:

So:

3. Relate temperature change to internal energy change For an ideal gas, the change in internal energy is:

Thus:

The root mean square (RMS) speed of gas molecules is a measure of the average speed of the molecules in a gas. It's not a simple average, but rather a statistical average that gives more weight to faster molecules, which is relevant for phenomena like diffusion and pressure. The kinetic theory of gases provides a direct relationship between the RMS speed, the absolute temperature of the gas, and its molar mass.

The intuition behind this relationship is that at higher temperatures, molecules have more kinetic energy, and thus move faster. Conversely, for a given kinetic energy, lighter molecules will move faster than heavier ones.

The formula for the RMS speed ($v_{rms}$) of gas molecules is:

To find the ratio of RMS speeds, we will apply this formula to both helium and oxygen, ensuring all units are consistent, especially temperature in Kelvin.

Here's how to solve the problem step-by-step:

1. Convert temperatures to absolute scale (Kelvin).

   The formula for RMS speed requires temperature in Kelvin. We convert the given Celsius temperatures by adding $273.15$ (or simply $273$ for exam purposes).

   • For helium: $T_{He} = 127^\circ C + 273 = 400 \, K$
   • For oxygen: $T_{O_2} = 527^\circ C + 273 = 800 \, K$
   Watch out

   A common mistake is to use Celsius temperatures directly in the RMS speed formula. Always convert to Kelvin!

2. Write down the RMS speed expressions for both gases.

   Using the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$:

   • For helium: $v_{rms, He} = \sqrt{\frac{3RT_{He}}{M_{He}}}$
   • For oxygen: $v_{rms, O_2} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$

   We are given the molar masses: $M_{He} = 4$ and $M_{O_2} = 32$. While molar mass is typically in $g/mol$, for a ratio, as long as both are in the same units (e.g., $g/mol$ or $kg/mol$), the units will cancel out.

3. Formulate the ratio of the RMS speeds.

   We need to find the ratio of RMS speeds of helium to oxygen: …

The key idea here is that internal energy of an ideal gas is purely kinetic — it comes from the random motion of molecules. For a diatomic gas like nitrogen ($N_2$), oxygen ($O_2$), and hydrogen ($H_2$), each molecule has 5 degrees of freedom at ordinary temperatures (3 translational + 2 rotational). The internal energy per mole is $\frac{f}{2}RT$, where $f$ is the number of degrees of freedom.

Since all three gases are diatomic and at the same temperature $T$, each mole contributes $\frac{5}{2}RT$ to the total internal energy. The mixture is just the sum of the contributions from each gas.

1. Identify the degrees of freedom.

   Nitrogen, oxygen, and hydrogen are all diatomic molecules. At temperatures where vibrational modes are not excited (which is the case in most standard problems unless stated otherwise), each has $f = 5$ degrees of freedom.

   Watch out

   A common mistake is to treat hydrogen as monatomic or to add vibrational degrees of freedom. Unless the problem explicitly mentions high temperature, stick with $f = 5$ for diatomic gases.

2. Write the internal energy per mole.

   For one mole of a diatomic ideal gas:

3. Calculate the contribution of each gas. …

Concept and intuition:

The molar specific heat at constant pressure, $C_p$, tells us how much heat is needed to raise the temperature of one mole of a gas by 1 K while allowing it to expand against a constant external pressure. For an ideal gas, the added heat does two things: it increases the internal energy (which depends only on temperature) and it does work as the gas expands. The key is that for a monoatomic gas, the internal energy comes only from translational kinetic energy, giving $C_v = \frac{3}{2}R$. Then, from the relation $C_p - C_v = R$ (true for any ideal gas), we get $C_p = \frac{5}{2}R$. This is a fundamental result — no need to memorize numbers, just the ratio of specific heats.

Step-by-step reasoning:

1. Recall the degrees of freedom for a monoatomic gas. A monoatomic gas (like helium or argon) has only three translational degrees of freedom. Each degree contributes $\frac{1}{2}R$ to the molar heat capacity at constant volume. So:

2. Use the universal relation between $C_p$ and $C_v$ for an ideal gas. For any ideal gas, the difference between the molar specific heats at constant pressure and constant volume is equal to the universal gas constant:

This comes from the first law and the ideal gas law: at constant pressure, the gas does work $P\Delta V = R\Delta T$, so extra heat is needed.

3. Substitute $C_v$ into the relation.

4. Plug in the given value of $R$. $R = 8.3\ \mathrm{J\,mol^{-1}\,K^{-1}}$, so: $$$$

Concept. The specific heat capacity of water (per gram) is one of the most well-known constants in physics and chemistry: $c = 1~\mathrm{cal/(g \cdot {}^\circ C)}$ (equivalently $\approx 4.186~\mathrm{J/(g\cdot K)}$). To get the molar specific heat, multiply by water's molar mass.

1. Molar mass of water: $M = 18~\mathrm{g/mol}$.

2. Molar heat capacity:

3. Express in terms of $R$. Using the commonly used approximation $R \approx 2~\mathrm{cal\,mol^{-1}K^{-1}}$:

This matches option (D). (This can also be checked in SI units: the real molar heat capacity of liquid water is about $75.3~\mathrm{J\,mol^{-1}K^{-1}}$, and $75.3 / 8.314 \approx 9.06$, confirming the same result.) …

When a diatomic gas dissociates at constant temperature, the heat supplied doesn't raise the average kinetic energy (temperature is fixed) — instead it goes into changing the internal energy associated with the gas's degrees of freedom as molecules break into atoms. A diatomic molecule (at moderate temperature) has 5 degrees of freedom (3 translational + 2 rotational), giving $U=\frac{5}{2}nRT$; a monatomic atom has only 3 translational degrees of freedom, giving $U=\frac{3}{2}nRT$.

Step-by-step solution

1. Initial state: 4 moles of diatomic gas at temperature $T$.

2. After dissociation: 2 moles of diatomic molecules break apart into $2\times2=4$ moles of monatomic atoms, leaving $4-2=2$ moles of diatomic gas undissociated. Total: 2 moles diatomic + 4 moles monatomic. $U_f = \left(2\times\frac{5}{2}RT\right) + \left(4\times\frac{3}{2}RT\right) = 5RT + 6RT = 11RT$ …

The key here is the first law of thermodynamics and the specific heat capacities of a diatomic gas. When heat is added at constant pressure, the gas expands and does work on the surroundings. The question asks for the work done on the gas — which is the negative of the work done by the gas. We need to connect the heat supplied to the work, using the fact that for an ideal gas, the ratio of heat capacities $\gamma = C_p/C_v$ is known for a diatomic gas.

1. First law and constant pressure process

   The first law states: $\Delta U = Q + W$, where $W$ is work done on the gas. For a constant pressure process, the work done by the gas is $W_{\text{by}} = P \Delta V$, so work done on the gas is $W = -P \Delta V$. The heat added at constant pressure is $Q = n C_p \Delta T$.

2. Relating heat to work via internal energy

   For an ideal gas, the change in internal energy depends only on temperature: $\Delta U = n C_v \Delta T$. Substituting into the first law:

So $W = n (C_v - C_p) \Delta T = -n R \Delta T$, since $C_p - C_v = R$.

Thus the work done on the gas is $W = -n R \Delta T$.

3. Finding $n \Delta T$ from the heat supplied We know $Q = n C_p \Delta T = 700$ J. For a diatomic gas, $C_v = \frac{5}{2}R$ and $C_p = \frac{7}{2}R$ (at moderate temperatures, ignoring vibrational modes). So:

4. Work done on the gas …

The problem asks for the value of $\alpha$ such that the heat exchange is zero — meaning the process is adiabatic. For an adiabatic process, the first law gives $dQ = 0$, so $dU = -dW$. The key is to connect the given relation $T^2 V^\alpha = \text{constant}$ to the standard adiabatic relation for an ideal gas.

For an ideal gas undergoing an adiabatic (reversible) process, we have $PV^\gamma = \text{constant}$, where $\gamma = C_P/C_V$. Using the ideal gas law $PV = nRT$, we can rewrite this in terms of $T$ and $V$:

So the standard adiabatic relation in $T$ and $V$ is $T V^{\gamma-1} = \text{constant}$.

Now, the given relation is $T^2 V^\alpha = \text{constant}$. For this to represent an adiabatic process, the exponent of $T$ must be 1 (as in the standard form), not 2. But wait — we can raise both sides of the standard relation to any power and it remains constant. For example, if $T V^{\gamma-1} = k$, then squaring gives $T^2 V^{2(\gamma-1)} = k^2$, which is also constant. So the given $T^2 V^\alpha = \text{constant}$ matches the squared form of the adiabatic relation if:

Now we need $\gamma$ for a polyatomic gas. A polyatomic gas (non-linear, at ordinary temperatures) has degrees of freedom $f = 6$ (3 translational + 3 rotational). Then:

Thus: …