To solve this problem, we need to determine the number of moles \(N\) of a polyatomic gas with degrees of freedom \(f = 6\) that must be mixed with two moles of a monoatomic gas (which has degrees of freedom \(f = 3\)) so that the mixture behaves like a diatomic gas (which has degrees of freedom \(f = 5\)).
The degrees of freedom \(f\) for a mixture of gases can be determined using the formula:
\(f_{\text{mix}} = \frac{N_1 \cdot f_1 + N_2 \cdot f_2}{N_1 + N_2}\)
where:
We are given:
Substituting these values into the formula:
\(5 = \frac{N \cdot 6 + 2 \cdot 3}{N + 2}\)
We solve for \(N\):
\(5(N + 2) = 6N + 6\)
Expanding both sides:
\(5N + 10 = 6N + 6\)
Rearranging the equation:
\(5N + 10 - 6 = 6N\)
\(5N + 4 = 6N\)
Simplify to solve for \(N\):
\(4 = 6N - 5N\)
\(N = 4\)
Thus, the value of \(N\) that satisfies this condition is 4. Therefore, the correct answer is 4.
The average degrees of freedom for the mixture, \( f_{\text{eq}} \), can be expressed as:
\(f_{\text{eq}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}\)
where:
- \( n_1 = N \) (number of moles of polyatomic gas),
- \( n_2 = 2 \) (number of moles of monoatomic gas),
- \( f_1 = 6 \) (degrees of freedom for polyatomic gas),
- \( f_2 = 3 \) (degrees of freedom for monoatomic gas),
- \( f_{\text{eq}} = 5 \) (degrees of freedom for diatomic gas).
Now, we substitute these values into the equation:
\(5 = \frac{N \times 6 + 2 \times 3}{N + 2}\)
Simplify the equation:
\(5 = \frac{6N + 6}{N + 2}\)
Multiply both sides by \( (N + 2) \):
\(5(N + 2) = 6N + 6\)
\(5N + 10 = 6N + 6\)
\(10 - 6 = 6N - 5N\)
\(N = 4\)
Thus, the value of \( N \) is 4.
The Correct Answer is: 4
An amount of ice of mass \( 10^{-3} \) kg and temperature \( -10^\circ C \) is transformed to vapor of temperature \( 110^\circ C \) by applying heat. The total amount of work required for this conversion is,
(Take, specific heat of ice = 2100 J kg\(^{-1}\) K\(^{-1}\),
specific heat of water = 4180 J kg\(^{-1}\) K\(^{-1}\),
specific heat of steam = 1920 J kg\(^{-1}\) K\(^{-1}\),
Latent heat of ice = \( 3.35 \times 10^5 \) J kg\(^{-1}\),
Latent heat of steam = \( 2.25 \times 10^6 \) J kg\(^{-1}\))
Match List-I with List-II.
Consider the sound wave travelling in ideal gases of $\mathrm{He}, \mathrm{CH}_{4}$, and $\mathrm{CO}_{2}$. All the gases have the same ratio $\frac{\mathrm{P}}{\rho}$, where P is the pressure and $\rho$ is the density. The ratio of the speed of sound through the gases $\mathrm{v}_{\mathrm{He}}: \mathrm{v}_{\mathrm{CH}_{4}}: \mathrm{v}_{\mathrm{CO}_{2}}$ is given by