Figure shows the variation in temperature ($$\Delta T$$) with the amount of heat supplied (Q) in an isobaric process corresponding to a monoatomic (M), diatomic (D) and a polyatomic (P) gas. The initial state of all the gases are the same and the scales for the two axes coincide. Ignoring vibrational degrees of freedom, the lines a, b and c respectively correspond to :

The problem involves identifying which lines a, b, and c correspond to monoatomic (M), diatomic (D), and polyatomic (P) gases on a graph of temperature change ($$\Delta T$$) versus heat supplied (Q) for an isobaric process. The initial state is the same for all gases, and vibrational degrees of freedom are ignored.

In an isobaric process, the heat supplied Q is related to the change in temperature $$\Delta T$$ by the equation:

$$ Q = n C_P \Delta T $$

where n is the number of moles and $$C_P$$ is the molar specific heat at constant pressure. Rearranging for $$\Delta T$$:

$$ \Delta T = \frac{Q}{n C_P} $$

Since n is the same for all gases (same initial state), $$\Delta T$$ is inversely proportional to $$C_P$$. Therefore, for a given Q, a higher $$C_P$$ results in a smaller $$\Delta T$$. On the graph with $$\Delta T$$ on the y-axis and Q on the x-axis, the slope of each line ($$\Delta T$$ per unit Q) is inversely proportional to $$C_P$$. Thus, a gas with a larger $$C_P$$ will have a flatter slope, and a gas with a smaller $$C_P$$ will have a steeper slope.

Now, we determine $$C_P$$ for each gas type, ignoring vibrational degrees of freedom:

• For a monoatomic gas (M), the degrees of freedom f = 3 (translational only). The molar specific heat at constant volume is $$C_V = \frac{f}{2} R = \frac{3}{2} R$$. Then, $$C_P = C_V + R = \frac{3}{2} R + R = \frac{5}{2} R$$.
• For a diatomic gas (D), f = 5 (3 translational + 2 rotational). So, $$C_V = \frac{5}{2} R$$, and $$C_P = \frac{5}{2} R + R = \frac{7}{2} R$$.
• For a polyatomic gas (P), we consider f = 6 (3 translational + 3 rotational, assuming non-linear structure, which is standard when vibrational modes are ignored). Thus, $$C_V = \frac{6}{2} R = 3R$$, and $$C_P = 3R + R = 4R$$.

Comparing the $$C_P$$ values:

$$ C_P(\text{M}) = \frac{5}{2}R = 2.5R $$

$$ C_P(\text{D}) = \frac{7}{2}R = 3.5R $$

$$ C_P(\text{P}) = 4R = 4.0R $$

So, $$C_P(\text{M}) < C_P(\text{D}) < C_P(\text{P})$$.

Since $$\Delta T$$ is inversely proportional to $$C_P$$, the slope of the line ($$\Delta T$$ vs Q) is largest for the gas with the smallest $$C_P$$ and smallest for the gas with the largest $$C_P$$:

• Monoatomic gas (M) has the smallest $$C_P$$ (2.5R), so it has the steepest slope.
• Diatomic gas (D) has a medium $$C_P$$ (3.5R), so it has a medium slope.
• Polyatomic gas (P) has the largest $$C_P$$ (4.0R), so it has the flattest slope.

Therefore, the lines on the graph, ordered from steepest to flattest slope, correspond to M, D, P. The problem asks for lines a, b, and c respectively. Given the correct answer is Option B, which is M, D, P, this means:

• Line a corresponds to monoatomic gas (M).
• Line b corresponds to diatomic gas (D).
• Line c corresponds to polyatomic gas (P).

Hence, the correct answer is Option B.