An empty LPG cylinder weighs 14.8 kg. When full, it weighs 29.0 kg and shows a pressure of 3.47 atm. In the course of use at ambient temperature, the mass of the cylinder is reduced to 23.0 kg. The final pressure inside the cylinder is _ _ _ _ atm. (Nearest integer)
(Assume LPG to be an ideal gas)

Solution

Weight of gas before use = 29.0 – 14.8 = 14.2 kg

Weight of gas after use = 23.0 – 14.8 = 8.2 kg

Now, $\frac {P_1}{n_1} = \frac {P_2}{n_2} $

$\Rightarrow \frac {P_1}{w_1} = \frac {P_2}{w_2} $

$\therefore \frac {3.47}{14.2} =  \frac {P_2}{8.2} $

$\Rightarrow P_2 = \frac {3.47 \times 8.2}{14.2} \approx 2 atm $

Answer: 2

The average translational kinetic energy of $N_2 $ gas molecules at _ _ _ _ $^\circ C $ becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 Volt. (Given $k_b = 1.38 \times 10^{-23} J/K $) (Fill the nearest integer)

Solution

K.E. of electron =  Translational K.E. of $N_2 $ molecules

$\therefore eV = \frac {3}{2} kT $

$\therefore 2 \times 1.6 \times 10^{-19} \times 0.1 = 3 \times 1.38 \times 10^{-23} \times T $

$\therefore T = \frac {1.6 \times 10^3 }{3 \times 0.69 } \approx 773 K = 500 ^\circ C $

Answer: 500