\(P_1V = n_1RT\)
\(P_2V = n_2RT\)
\(⇒ (100 kPa) V = (n_1 + n_2)RT\)
\(⇒n_1+n_2=\frac{(100 kPa)(2000 cm)^3)}{8.3×300}….(1)\)
Also, \(n_1 × 2 + n_2 × 32 = 0.76 ….(2)\)
Solving (1) and (2),
\(n_1 = 0.06\)
\(n_2 = 0.02\)
\(⇒ \frac{n_1}{n_2} = 3\)
Hence, the correct option is (B): \(\frac{3}{1}\)
The temperature at which the rms speed of oxygen molecules is 75\% of the rms speed of nitrogen molecules at a temperature of \( 287^\circ C \) is:


In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
The gas laws were developed at the end of the 18th century, when scientists began to realize that relationships between pressure, volume and temperature of a sample of gas could be obtained which would hold to approximation for all gases.