Question

$1$ mole of $H_2$ gas is contained in a box of volume $V=1.00$ $m^3$ at $T=300$K. The gas is heated to a temperature of $T=3000$K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be(considering all gases to be ideal).

• A. Same as the pressure initially
• B. $2$ times the pressure initially
• C. $10$ times the pressure initially
• D. $20$times the pressure initially

Answer 1

Answer: (D)

$20$times the pressure initially

According to gas equation,
$PV=nRT$
or $\frac{P_2{V}_2}{T_2}=\frac{P_1{V}_1}{T_1}$ or $\frac{P_2}{P_1}=\frac{V_1}{V_2}\cdot \frac{T_2}{T_1}$
Here, $T_2=3000$K, $T_1=300$K
Since $H_2$ splits into hydrogen atoms, therefore volume become half i.e., $V_2=\frac{1}{2}{V}_1$
$\therefore \frac{P_2}{P_1}=\frac{V_1}{\frac{1}{2}{V}_1}\times \frac{3000}{300}$ or $\frac{P_2}{P_1}=20$