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

Core Logic

Given energy is in electron-volts (1 eV = 1.6 × 10⁻¹⁹ J$1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$), we isolate T$T$:

Substitute constants (kB = 1.38 × 10⁻²³ J/K$k_B = 1.38 \times 10^{-23}\text{ J/K}$):

Step 1: Compute value

Pattern Recognition

Converting eV energy properties straight to structural SI standard Joules reveals highly cleanly simplified scalar components when paired with Boltzmann values.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

Related Formula

From the Ideal Gas Law equation:

Core Logic

Given that both vessels possess the same volume (VA = VB$V_A = V_B$) and identical temperature states (TA = TB$T_A = T_B$), the ratio of pressure reduces directly to:

where nA$n_A$ and nB$n_B$ are the number of moles of Hydrogen and Oxygen respectively.

Step 1: Calculate the Number of Moles

For Hydrogen (H₂$H_2$, molar mass = 2 ~g/mol$2 \mathrm{~g/mol}$):

For Oxygen (O₂$O_2$, molar mass = 32 ~g/mol$32 \mathrm{~g/mol}$):

Step 2: Find the Pressure Ratio

Substituting mole counts into the direct ratio:

Therefore, the pressure ratio (PA)/(PB)$\frac{P_A}{P_B}$ is 16$16$.

Pattern Recognition

For gas mixtures or vessel comparisons under constant volume and temperature, pressure matches the molar abundance directly (P ∝ n$P \propto n$). Remember that standard elementary gases (H₂, O₂, N₂$H_2, O_2, N_2$) exist as diatomic configurations when defining molar values.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

Related Formula

Core Logic

For a monoatomic gas, degrees of freedom f₁ = 3$f_1 = 3$. Moles n₁ = 3$n_1 = 3$. For a diatomic gas, degrees of freedom f₂ = 5$f_2 = 5$. Moles n₂ = 2$n_2 = 2$. We can compute the equivalent degrees of freedom for the mixture using a weighted average.

Step 1: Calculate Equivalent Degrees of Freedom

Step 2: Calculate Adiabatic Exponent

Pattern Recognition

Alternatively, you can compute Cv$C_v$ and Cₚ$C_p$ for the mixture: Cv,mix = n₁ Cv1 + n₂ Cv2n₁ + n₂$C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$, and γmix = Cp,mixCv,mix$\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}}$. Both methods yield identical results rapidly.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

Related Formula

Core Logic

For the RMS velocities to be equal, the ratio of temperature to molar mass (T/M$T/M$) must be identical for both gases.

Step 1: Set Up Equivalency

Step 2: Substitute Values

TO₂ = 47^ = 47 + 273 = 320 ~K$T_{O_2} = 47^\circ\mathrm{C} = 47 + 273 = 320 \mathrm{~K}$ MH₂ = 2 ~g/mol$M_{H_2} = 2 \mathrm{~g/mol}$ MO₂ = 32 ~g/mol$M_{O_2} = 32 \mathrm{~g/mol}$

Pattern Recognition

vrms$v_{\text{rms}}$ scales strictly as √(T/M)$\sqrt{T/M}$. Remember to always convert Celsius to Kelvin before substituting.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

Related Formula

U = n Cv T$U = n C_v T$ where Cv = (f)/(2) R$C_v = \frac{f}{2} R$ and f$f$ is the degree of freedom.

Core Logic

Argon (Ar) is monatomic f₁ = 3 Cv₁ = (3R)/(2)$\implies f_1 = 3 \implies C_{v_1} = \frac{3R}{2}$. Oxygen (O₂$O_2$) is diatomic f₂ = 5$\implies f_2 = 5$ (neglecting vibrational modes) Cv₂ = (5R)/(2)$\implies C_{v_2} = \frac{5R}{2}$. Total internal energy U = U₁ + U₂ = n₁ Cv₁ T + n₂ Cv₂ T$U = U_1 + U_2 = n_1 C_{v_1} T + n_2 C_{v_2} T$.

Step 1: Compute Total Energy

U = 12RT + 15RT$U = 12RT + 15RT$ U = 27 RT$U = 27 RT$

Pattern Recognition

Internal energy is strictly additive. Immediately map Monatomic → 3/2$\to 3/2$ and Diatomic → 5/2$\to 5/2$. Plug linearly: 8(1.5) + 6(2.5) = 12 + 15 = 27$8(1.5) + 6(2.5) = 12 + 15 = 27$.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics