There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K, at steady state the pressure in the vessels will be (in kPa).

### Related Formula fracP_1T_1 = fracP_2T_2$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$ ### Core Logic Initial temperature T_1 = 27 + 273 = 300text K$T_1 = 27 + 273 = 300\text{ K}$. Since volume is kept fixed : fracP300 = frac2PT_2 implies T_2 = 600text K$$\frac{P}{300} = \frac{2P}{T_2} \implies T_2 = 600\text{ K}$$ Converting back to Celsius : T_2 = 600 - 273 = 327^circmathrmC$$T_2 = 600 - 273 = 327^{\circ}\mathrm{C}$$ ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula Molar specific heat at constant volume for a gas mixture: C_Vtext, mix = fracn_1 C_V1 + n_2 C_V2n_1 + n_2$$C_{V\text{, mix}} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$$ For a monoatomic gas: C_V1 = frac32R$$C_{V1} = \frac{3}{2}R$$ For a diatomic gas: C_V2 = frac52R$$C_{V2} = \frac{5}{2}R$$ ### Core Logic Given values: n_1 = 2$n_1 = 2$ (monoatomic), n_2 = 6$n_2 = 6$ (diatomic). Substitute these inputs directly into the mixture equation: C_Vtext, mix = frac2 times left(frac32Rright) + 6 times left(frac52Rright)2 + 6$$C_{V\text{, mix}} = \frac{2 \times \left(\frac{3}{2}R\right) + 6 \times \left(\frac{5}{2}R\right)}{2 + 6}$$ ### Step 1: Simplify Expression C_Vtext, mix = frac3R + 15R8 = frac18R8 = frac94R$$C_{V\text{, mix}} = \frac{3R + 15R}{8} = \frac{18R}{8} = \frac{9}{4}R$$ ### Pattern Recognition Weighted average rule based on internal degrees of freedom: total internal energy changes scale additively with mole numbers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

### Related Formula The state equation of an ideal gas in terms of the number of molecules N$N$ and Boltzmann constant k$k$ is: PV = NkT$PV = NkT$ Rearranging to express pressure in terms of number density n = N/V$n = N/V$: P = n k T$P = n k T$ ### Core Logic Given parameters: * Number density, n = fracNV = 2.0 times 10^25text molecules/m^3$n = \frac{N}{V} = 2.0 \times 10^{25}\text{ molecules/m}^3$ * Pressure, P = 1.38text atm = 1.38 times 1.01 times 10^5text N/m^2$P = 1.38\text{ atm} = 1.38 \times 1.01 \times 10^5\text{ N/m}^2$ * Boltzmann constant, k = 1.38 times 10^-23text J K^-1$k = 1.38 \times 10^{-23}\text{ J K}^{-1}$ ### Step 1: Solve for Temperature Rearranging P = n k T$P = n k T$ for temperature T$T$: T = fracPnk$$T = \frac{P}{nk}$$ Substitute the values: T = frac1.38 times 1.01 times 10^5(2.0 times 10^25) times (1.38 times 10^-23)$$T = \frac{1.38 \times 1.01 \times 10^5}{(2.0 \times 10^{25}) \times (1.38 \times 10^{-23})}$$ Notice that the term 1.38$1.38$ cancels out from numerator and denominator: T = frac1.01 times 10^52.0 times 10^2$$T = \frac{1.01 \times 10^5}{2.0 \times 10^2}$$ T = frac1.01 times 10^32.0 approx frac10102 approx 500text K$$T = \frac{1.01 \times 10^3}{2.0} \approx \frac{1010}{2} \approx 500\text{ K}$$ ### Pattern Recognition The numerical values are designed to cancel out smoothly. Spotting the 1.38$1.38$ cancelation instantly saves valuable calculation time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula The equivalent degrees of freedom f_texteq$f_{\text{eq}}$ for a mixture of gases is: f_texteq = fracn_1 f_1 + n_2 f_2n_1 + n_2$$f_{\text{eq}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$$ where: * n_1, n_2$n_1, n_2$ are the number of moles of each gas. * f_1, f_2$f_1, f_2$ are the respective degrees of freedom of each gas. ### Core Logic For the given gases: 1. Polyatomic gas: * Moles, n_1 = N$n_1 = N$ * Degrees of freedom, f_1 = 6$f_1 = 6$ 2. Monoatomic gas: * Moles, n_2 = 2$n_2 = 2$ * Degrees of freedom, f_2 = 3$f_2 = 3$ We want the mixture to behave as a diatomic gas. For a diatomic gas: * Equivalent degrees of freedom, f_texteq = 5$f_{\text{eq}} = 5$ ### Step 1: Solve for N Substitute the values into the degrees of freedom mixture formula: 5 = frac(N)(6) + (2)(3)N + 2$$5 = \frac{(N)(6) + (2)(3)}{N + 2}$$ 5(N + 2) = 6N + 6$$5(N + 2) = 6N + 6$$ 5N + 10 = 6N + 6$$5N + 10 = 6N + 6$$ 10 - 6 = 6N - 5N implies N = 4$$10 - 6 = 6N - 5N \implies N = 4$$ ### Pattern Recognition Diatomic equivalent degree of freedom is 5. Since the monoatomic degrees of freedom (3) and polyatomic degrees of freedom (6) bracket 5, you can use the weighted ratio method to find the molar proportions directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula K_textavg = frac32 k_B T$$K_{\text{avg}} = \frac{3}{2} k_B T$$ ### Core Logic Given energy is in electron-volts (1text eV = 1.6 times 10^-19text J$1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$), we isolate T$T$: T = frac2 K_textavg3 k_B$$T = \frac{2 K_{\text{avg}}}{3 k_B}$$ Substitute constants (k_B = 1.38 times 10^-23text J/K$k_B = 1.38 \times 10^{-23}\text{ J/K}$): ### Step 1: Compute value T = frac2 times 0.414 times 1.6 times 10^-193 times 1.38 times 10^-23$$T = \frac{2 \times 0.414 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}}$$ T = frac1.3248 times 10^-194.14 times 10^-23 = 0.32 times 10^4 = 3200text K$$T = \frac{1.3248 \times 10^{-19}}{4.14 \times 10^{-23}} = 0.32 \times 10^4 = 3200\text{ K}$$ ### Pattern Recognition Converting eV energy properties straight to structural SI standard Joules reveals highly cleanly simplified scalar components when paired with Boltzmann values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases