To find the ratio of the specific heats at constant volume for a mixture of gases, we need to calculate the individual specific heats first.
Let's start with some fundamental concepts:
Given one mole of a monoatomic gas and one mole of a diatomic gas, the total number of moles \( n = 2 \).
The average specific heat at constant volume for the mixture, \( C_{V_{\text{mix}}} \), can be calculated as the weighted average:
\(C_{V_{\text{mix}}} = \frac{(1 \cdot C_{V_\text{monoatomic}}) + (1 \cdot C_{V_\text{diatomic}})}{2}\)
Substituting the values:
\(C_{V_{\text{mix}}} = \frac{(1 \cdot \frac{3}{2}R) + (1 \cdot \frac{5}{2}R)}{2}\) \(C_{V_{\text{mix}}} = \frac{\frac{3}{2}R + \frac{5}{2}R}{2}\) \(C_{V_{\text{mix}}} = \frac{8}{4}R = 2R\)
Now let's find the ratio of the specific heats for the gases:
\(\text{Ratio} = \frac{C_{V_\text{monoatomic}}}{C_{V_\text{mix}}} = \frac{\frac{3}{2}R}{2R}\) \(\text{Ratio} = \frac{3}{4}\)
Thus, the correct option was expected to be \(\frac{3}{4}\).
However, the given correct answer is \(\frac{3}{5}\), which seems incorrect based on our calculation.
The specific heat capacities at constant volume are:
\[ (C_V)_{\text{mono}} = \frac{3}{2}R, \quad (C_V)_{\text{dia}} = \frac{5}{2}R. \]
The ratio is:
\[ \frac{(C_V)_{\text{mono}}}{(C_V)_{\text{dia}}} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5}. \]
Final Answer: \(3 : 5\).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,



What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)