Two vessels A and B are of the same size and are at the same temperature. A contains 1 g of hydrogen and B contains 1 g of oxygen. \(P_A\) and \(P_B\) are the pressures of the gases in A and B respectively, then \(\frac{P_A}{P_B}\) is:
16
8
4
32
Step 1: Use the Ideal Gas Equation:
\[ \frac{P_A V_A}{P_B V_B} = \frac{n_A R T_A}{n_B R T_B} \]
- Given \(V_A = V_B\) and \(T_A = T_B\), the equation simplifies to:
\[ \frac{P_A}{P_B} = \frac{n_A}{n_B} \]
Step 2: Calculate Moles of Each Gas:
- For hydrogen in vessel A:
\[ n_A = \frac{\text{mass of hydrogen}}{\text{molar mass of } H_2} = \frac{1 \text{ g}}{2 \text{ g/mol}} = \frac{1}{2} \text{ mol} \]
- For oxygen in vessel B:
\[ n_B = \frac{\text{mass of oxygen}}{\text{molar mass of } O_2} = \frac{1 \text{ g}}{32 \text{ g/mol}} = \frac{1}{32} \text{ mol} \]
Step 3: Calculate the Ratio of Pressures:
\[ \frac{P_A}{P_B} = \frac{n_A}{n_B} = \frac{\frac{1}{2}}{\frac{1}{32}} = \frac{1}{2} \times 32 = 16 \]
So, the correct answer is: 16
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is:
Pressure is defined as the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
When a force of ‘F’ Newton is applied perpendicularly to a surface area ‘A’, then the pressure exerted on the surface by the force is equal to the ratio of F to A. The formula for pressure (P) is:
P = F / A
The SI unit of pressure is the pascal (Pa)
A pascal can be defined as a force of one newton applied over a surface area of a one-meter square.