Categories
Physics Olympiad

Specific Heat of a Rigid Triangular Molecule

A rigid triangular molecule consists of three non-collinear atoms joined by rigid rods. The constant pressure molar specific heat (C_p) of an ideal gas consisting of such molecules is

(a) (6R)

(b) (5R)

(c) (4R)

(d) (3R)

Degrees of freedom are the number of independent parameters that define its configuration
If (N) be the number of particles in a system and (k) be the number of constraints between the number of degrees of freedom is given by $$ f=3N-k$$ $$f=(3*3)-3$$ $$=6$$
Relation between (f) and (C_p) $$ C_p=(f/2+1)R$$ $$ \Rightarrow C_p=(6/2+1)R$$ $$C_p=4R$$

Categories
Physics Olympiad

Variation of Specific Heat

In this post, let’s learn about variation of specific heat by finding out the difference between mean specific heat and specific heat at midpoint.

The Problem:

The variation of the specific heat of a substance is given by the expression $$ C=A+BT^2$$ where (A) and (B) are constants and (T) is the Celsius temperature. Find the difference between the mean specific heat and specific heat at midpoint.

Discussion:

The variation of the specific heat of a substance is given by the expression $$ C=A+BT^2$$ where (A) and (B) are constants and (T) is the Celsius temperature.

Mean specific heat
$$ \bar{C}=\frac{\int C dT}{dT}=\frac{\int_{0}^{T}(A+BT^2)dT}{T}$$ $$= \frac{AT+BT^3/3}{T}$$ $$=A+BT^2/2$$
C(midpoint)$$ = A+B(T/2)^2$$ $$=A+\frac{BT^2}{4}$$
Hence, the difference between mean specific heat and specific heat at midpoint $$= \bar{C}-C(midpoint)$$ $$=A+BT^2/3-(A+BT^2/4)$$ $$=\frac{BT^2}{12}$$