The compressibility factor \( Z \) is given by: \[ Z = \frac{V_{\text{real}}}{V_{\text{ideal}}} = 0.5 \]
The real volume \( V_{\text{real}} \) is: \[ V_{\text{real}} = 0.4 \, \text{dm}^3 \, \text{mol}^{-1} = 0.4 \, \text{L/mol} \]
Using the equation for ideal volume: \[ V_{\text{ideal}} = \frac{0.4}{0.5} = 0.8 \, \text{L/mol} \]
Therefore, \( y = 0.8 \, \text{L/mol} \).
Using the ideal gas law, \( PV = nRT \), we calculate the pressure \( P \): \[ P = \frac{1 \times 8 \times 10^{-2} \times 800}{0.8} \]
Solving for \( P \), we get \( P = 80 \, \text{atm} \).
Finally, the ratio \( \frac{x}{y} \) is: \[ \frac{x}{y} = \frac{80}{0.8} = 100 \]
The pH of a 0.01 M weak acid $\mathrm{HX}\left(\mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}\right)$ is found to be 5 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6 . The new concentration of the diluted weak acid is given as $\mathrm{x} \times 10^{-4} \mathrm{M}$. The value of x is _______ (nearest integer).
A body of mass $m$ is suspended by two strings making angles $\theta_{1}$ and $\theta_{2}$ with the horizontal ceiling with tensions $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ simultaneously. $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ are related by $\mathrm{T}_{1}=\sqrt{3} \mathrm{~T}_{2}$. the angles $\theta_{1}$ and $\theta_{2}$ are
Consider the following equilibrium,
CO(g) + 2H2(g) ↔ CH3OH(g)
0.1 mol of CO along with a catalyst is present in a 2 dm3 flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH3OH is formed. The Kp is ____ × 10-3 (nearest integer).
Given: R = 0.08 dm3 bar K-1mol-1
Assume only methanol is formed as the product and the system follows ideal gas behaviour.
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____. 
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
As shown in the figures, a uniform rod $ OO' $ of length $ l $ is hinged at the point $ O $ and held in place vertically between two walls using two massless springs of the same spring constant. The springs are connected at the midpoint and at the top-end $ (O') $ of the rod, as shown in Fig. 1, and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $ f_1 $. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2, and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $ f_2 $. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:
The equilibrium constant may be defined as the ratio between the product of the molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration term raised to a power equal to the stoichiometric coefficient in the balanced chemical reaction.
The equilibrium constant at a given temperature is the ratio of the rate constant of forwarding and backward reactions.
Kequ = kf/kb = [C]c [D]d/[A]a [B]b = Kc
where Kc, indicates the equilibrium constant measured in moles per litre.
For reactions involving gases: The equilibrium constant formula, in terms of partial pressure will be:
Kequ = kf/kb = [[pC]c [pD]d]/[[pA]a [pB]b] = Kp
Where Kp indicates the equilibrium constant formula in terms of partial pressures.
Medium Kc/Kp values indicate optimum product formation.
The equilibrium constant is the ratio of the concentrations raised to the stoichiometric coefficients. Therefore, the unit of the equilibrium constant = [Mole L-1]△n.
where, ∆n = sum of stoichiometric coefficients of products – a sum of stoichiometric coefficients of reactants.