Question

H₂ gas is kept inside a container A and container B each having volume 2 litre under different c which are described below. Determining the missing values with proper unit. [R = 8 J mol⁻¹ K⁻¹ and Nᴀ = 6 x 10²³, N = No. of molecules]

Answer 1

(i) $1200 \, \text{Pa}$, (ii) $2.533 \times 10^{22}$, (iii) $3600 \, \text{J}$, (iv) $303.975 \, \text{J}$, (v) $1897.37 \, \text{m/s}$, (vi) $1549.19 \, \text{m/s}$

Answer 2

1. Calculated moles for Container A ($n_A$) from given molecules ($N_A$) and Avogadro's number ($N_A$): $n_A = \frac{6 \times 10^{20}}{6 \times 10^{23}} = 1 \times 10^{-3} \, \text{mol}$.
2. Used Ideal Gas Law ($PV=nRT$) to find Pressure for Container A ($P_A$): $P_A = \frac{n_A R T_A}{V} = \frac{(1 \times 10^{-3}) \times 8 \times 300}{2 \times 10^{-3}} = 1200 \, \text{Pa}$. This is (i).
3. Calculated Total Average Kinetic Energy for Container A ($KE_{avg, A}$) using $KE_{avg} = \frac{3}{2}nRT$: $KE_{avg, A} = \frac{3}{2} \times (1 \times 10^{-3}) \times 8 \times 300 = 3600 \, \text{J}$. This is (iii).
4. Calculated Root Mean Square Speed for Container A ($U_{rms, A}$) using $U_{rms} = \sqrt{\frac{3RT}{M}}$ (Molar mass of H₂ = 2 g/mol = $2 \times 10^{-3}$ kg/mol): $U_{rms, A} = \sqrt{\frac{3 \times 8 \times 300}{2 \times 10^{-3}}} = \sqrt{3.6 \times 10^6} \approx 1897.37 \, \text{m/s}$. This is (v).
5. Calculated Most Probable Speed for Container A ($Z_{1, A}$) using $Z_1 = \sqrt{\frac{2RT}{M}}$: $Z_{1, A} = \sqrt{\frac{2 \times 8 \times 300}{2 \times 10^{-3}}} = \sqrt{2.4 \times 10^6} \approx 1549.19 \, \text{m/s}$. This is (vi).
6. Calculated moles for Container B ($n_B$) from given pressure ($P_B = 1 \, \text{atm} = 101325 \, \text{Pa}$), volume ($V = 2 \times 10^{-3} \, \text{m}^3$), and temperature ($T_B = 600 \, \text{K}$) using Ideal Gas Law: $n_B = \frac{P_B V}{R T_B} = \frac{101325 \times 2 \times 10^{-3}}{8 \times 600} \approx 0.0422 \, \text{mol}$.
7. Calculated Number of Molecules for Container B ($N_B$) from moles ($n_B$) and Avogadro's number ($N_A$): $N_B = n_B \times N_A = 0.0422 \times 6 \times 10^{23} \approx 2.533 \times 10^{22}$. This is (ii).
8. Calculated Total Average Kinetic Energy for Container B ($KE_{avg, B}$) using $KE_{avg} = \frac{3}{2}nRT$: $KE_{avg, B} = \frac{3}{2} \times 0.0422 \times 8 \times 600 \approx 303.975 \, \text{J}$. This is (iv).
9. The value "0.7" in the table for Container B under "Ratio U_rms" is interpreted as the ratio $\frac{U_{rms, A}}{U_{rms, B}}$, as $\sqrt{\frac{T_A}{T_B}} = \sqrt{\frac{300}{600}} = \sqrt{0.5} \approx 0.707$. The direct calculation of $U_{rms, B}$ would be $\sqrt{\frac{3 \times 8 \times 600}{2 \times 10^{-3}}} \approx 2683.3 \, \text{m/s}$, which is not represented by 0.7.