Question

1 litre of a sample of $H_2$ is at $0^{o}C$ and 1 $atm$ pressure. 1 litre of another sample of hydrogen is at $0^{o}C$ and 2 $atm$ pressure. Kinetic energies of the gases are in the ratio of:

• A. 1 : 2
• B. 2 : 1
• C. 3: 1
• D. 1 : 1

Answer 1

Answer: (A)

1 : 2

We know

$p=\frac{2}{3}K.E$ So, $K.E=\frac{3}{2}p$

$(K.E{)}_{H_2}=$ kinetic energy of $H_2$ sample

$p_{H_2}$=$pressureof{H}_{2}gas$

$(K.E{)}_A=$ kinetic energy of another gas sample of hydrogen

$p_A=pressureofanothergassample$

$\frac{(K.E{)}_{H_2}}{(K.E{)}_A}=\frac{p_{H_2}}{p_A}=\frac{1}{2}$

Hence, option $A$ is correct.