For 1 mol of gas, the plot of pV vs. p is shown below. p is the pressure and V is the volume of the gas. What is the value of compressibility factor at point A?
When solving compressibility factor problems:
Start by expressing \(Z = \frac{PV}{RT}\) and substitute the appropriate gas law (e.g., Van der Waals equation).
Expand and simplify using the given conditions to isolate terms involving \(V\), \(a\), and \(b\).
Pay attention to the behavior of \(Z\) in real gas scenarios.
\(1-\frac{a}{RTV}\)
\(1+\frac{b}{V}\)
\(1-\frac{b}{V}\)
\(1+\frac{a}{RTV}\)
For 1 mole of real gas, \[ PV = ZRT \] From the graph, \(PV\) for a real gas is less than \(PV\) for an ideal gas at point A. Thus, \(Z < 1\). The compressibility factor \(Z\) is given by: \[ Z = 1 - \frac{a}{V_m RT} \] Substituting into the definition of \(Z\) and simplifying: \[ Z = 1 - \frac{a}{RTV} \]
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
If a substance ‘A’ dissolves in a solution of a mixture of ‘B’ and ‘C’ with their respective number of moles as \(n_a\), \(n_b\), and \(n_c\), the mole fraction of C in the solution is:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,