Degrees of Freedom for Gas Molecules | Problem and Solution
When a gas expands adiabatically, its volume is doubled while its absolute temperature is decreased by a factor (1.32). Compute the number of degrees of freedom for the gas molecules.
Solution:
The number of degrees can be found from the relation $$ f=\frac{2}{\gamma-1}$$
We can find (\gamma) from the adiabatic relation,$$ T_2V_2^{\gamma-1}= T_1V_1^{\gamma-1} $$
$$( \frac{V_2}{V_1})^{\gamma-1}=\frac{T_1}{T_2}=1.32$$
$$ 2^{\gamma-1}=1.32$$
where $$ \gamma=1+\frac{log 1.32}{log2}=1.4$$
The number of degrees of freedom $$ f=\frac{2}{1.4-1}=5$$
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) $$ C_p=(f/2+1)R$$ $$ \Rightarrow C_p=(6/2+1)R$$ $$C_p=4R$$
Work Done on Compression of Gas
A cylinder contains (16g) of (O_2). The work done when the gas is compressed to (75\%) of the original volume at constant temperature of (27^\circ) is ________.
Discussion:
Given mass of (O_2), m=(16g)
Number of moles of (O_2), $$n=\frac{m}{M}$$ where (M)=molecular weight of (O_2)=(32g)
$$n=\frac{16}{32}=\frac{1}{2}$$
Temperature (T=27^\circ=300K)
If (V_1) be the original volume and (V_2) be the final volume
Work done by the gas in the isothermal process $$ W_0=nRTlog(V_2/V_1)$$$$=0.58.31ln(3/4)$$$$=1508.31ln(3/4)$$$$=-358.56J$$
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 $$= \bar{C}-C(midpoint)$$ $$=A+BT^2/3-(A+BT^2/4)$$ $$=\frac{BT^2}{12}$$