Q.What amount of heat must be supplied to $2.0 \times 10^{-2}\ \text{kg}$ of nitrogen (at room temperature) to raise its temperature by $45\,^{\circ}\text{C}$ at constant pressure? (Molecular mass of $\text{N}_2 = 28$; $R = 8.3\ \text{J mol}^{-1}\ \text{K}^{-1}$.)

Applying the first law at constant pressure

Heat supplied at constant pressure:

Given $C_p = \dfrac{5R}{2}$, $Q = 1662\,\text{J}$ and $\Delta T = 50\,\text{K}$, with $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$:

Solving for the number of moles:

Step 1 — The set-up (from the figure)

Three identical rods meet at a common junction $x$:

• One rod runs to an end held at $\mathbf{0^\circ C}$ (the cold sink).
• Two rods run to ends held at $\mathbf{90^\circ C}$ (the hot sources).

The rods are of the same material and identical in all respects, so each has the same thermal conductivity $k$, the same cross-sectional area $A$ and the same length $L$. No heat escapes through the curved surfaces, so all the heat that arrives at the junction must leave through the rods.

Step 2 — The concept: steady-state conduction

The rate of heat conduction along a rod (Fourier's law) is

Define the thermal conductance $C = \dfrac{kA}{L}$, which is identical for all three rods. Then simply

In the steady state the junction's temperature is no longer changing, which means it is storing no heat. Therefore

This is the thermal analogue of Kirchhoff's junction rule for currents.

Step 3 — Write the heat currents

The junction is at temperature $x$, with $0 < x < 90$ (it must lie between the extremes).

Heat IN — along the two rods from the $90^\circ$C ends (heat flows hot → cold, so into the junction):

Heat OUT — along the one rod to the $0^\circ$C end:

Step 4 — Equate and solve

The conductance $C = kA/L$ is the same on both sides, so it cancels entirely — which is why we never needed the values of $k$, $A$ or $L$:

Step 5 — Why 60 °C and not the naïve 45 °C

The tempting wrong answer is $45^\circ$C — the plain average of $0$ and $90$. That would be right if one rod came from the hot end and one went to the cold end (a symmetric arrangement).

But here two rods feed heat in and only one carries it away. The junction is fed twice as strongly as it is drained, so it must settle closer to the hot end than to the cold one. Indeed $60^\circ$C is two-thirds of the way from $0$ to $90$ — exactly the weighted average:

Verify the balance: in-flow $= 2C(90-60) = 60C$; out-flow $= C(60) = 60C$. Equal ✓

ANSWER: A

Step 1 — The concept.

Work done by a gas is $W = \int P\,dV$, which is the area under the path on a $P$–$V$ diagram. Around a closed cycle the areas under the outgoing and returning legs partially cancel, leaving

positive if the cycle runs clockwise (the gas expands at high pressure and is compressed at low pressure — a net output, as in an engine) and negative if anticlockwise. Note also that for a full cycle $\Delta U = 0$ (internal energy is a state function), so by the first law $Q = W$ — but we do not need that here.

Step 2 — Read the vertices.

The three legs are: M→N a straight sloping line, N→O horizontal at $P = P_0$ (compression), O→M vertical at $V = V_0$ (isochoric pressure rise).

Step 3 — Compute the enclosed area.

The triangle is right-angled at $O$:

• horizontal leg $ON = 3V_0 - V_0 = 2V_0$,
• vertical leg $OM = 3P_0 - P_0 = 2P_0$.

Step 4 — Fix the sign from the sense of traversal.

Apply the shoelace formula to $M(1,3) \to N(3,1) \to O(1,1)$ (in units of $V_0$ and $P_0$):

A negative shoelace sum means the vertices are listed clockwise, and $|{-4}|/2 = 2$ ⇒ area $= 2P_0V_0$. Clockwise ⇒ the gas does positive net work.

(Physically: on M→N the gas expands from $V_0$ to $3V_0$ at relatively high pressures — large positive work; on N→O it is compressed back at the lowest pressure $P_0$ — a smaller negative work; on O→M the volume is constant so $W = 0$. The expansion wins, hence net positive work.)

Step 5 — Cross-check leg by leg.

• $W_{MN}$ = area of the trapezium under the straight line from $V_0$ to $3V_0$ $= \tfrac12(3P_0 + P_0)(3V_0 - V_0) = \tfrac12(4P_0)(2V_0) = 4P_0V_0$.
• $W_{NO} = P_0\,(V_0 - 3V_0) = -2P_0V_0$ (compression at constant $P_0$).
• $W_{OM} = 0$ (isochoric, $dV = 0$).

(The "monoatomic, one mole" data are not needed for the work — they would matter only for $Q$ or $\Delta U$ on individual legs.)

ANSWER: D

The key here is to connect the heat supplied at constant pressure to the two places it can go: increasing the internal energy of the gas, and doing work on the surroundings. The fraction that becomes work is simply the work done divided by the heat supplied. Since the process is at constant pressure, both the work and the heat have straightforward expressions in terms of the temperature change, and the ratio depends only on the gas's specific heat capacities — which for a monoatomic gas are fixed numbers.

Let’s walk through it step by step.

1. Identify the relevant thermodynamic quantities. For a monoatomic ideal gas, the molar specific heat at constant volume is $C_V = \frac{3}{2}R$, and at constant pressure it is $C_P = C_V + R = \frac{5}{2}R$. When heat $Q$ is supplied at constant pressure to $n$ moles, causing a temperature rise $\Delta T$, we have:

2. Find the work done during the expansion. At constant pressure $P$, the work done by the gas is $W = P \Delta V$. Using the ideal gas law $PV = nRT$, a change in volume at constant pressure gives $P \Delta V = nR \Delta T$. So:

3. Compute the fraction of heat converted to work. The fraction is $W/Q$:

The key here is to recognise that the gas is oxygen ($O_2$), which is diatomic. At constant pressure, the heat added equals the change in enthalpy: $Q = n C_p \Delta T$. And because pressure is constant, volume is directly proportional to absolute temperature (Charles’s law). So doubling the volume means doubling the absolute temperature.

Let’s walk through it step by step.

1. Identify the gas and its degrees of freedom.

   Oxygen ($O_2$) is a diatomic molecule. At moderate temperatures (like $27^\circ\text{C}$), it has 5 degrees of freedom: 3 translational and 2 rotational. Vibrational modes are not excited at this temperature.

   For a diatomic ideal gas:

   • $C_v = \frac{5}{2}R$
   • $C_p = C_v + R = \frac{7}{2}R$

2. Convert the initial temperature to Kelvin.

   $T_1 = 27^\circ\text{C} = 300\,\text{K}$

   Always use absolute temperature in gas law calculations.

3. Relate volume change to temperature change at constant pressure.

   For an ideal gas at constant pressure: $V \propto T$ (Charles’s law).

   So if volume doubles: $\frac{V_2}{V_1} = 2 \implies \frac{T_2}{T_1} = 2$

   Therefore $T_2 = 2 \times 300 = 600\,\text{K}$, and $\Delta T = 300\,\text{K}$.

4. Heat added at constant pressure.

   The heat required is $Q = n C_p \Delta T$.

   Here $n = 1$ mole, $C_p = \frac{7}{2}R$, $\Delta T = 300\,\text{K}$.

   So: