1 mole of H₂ gas is contained in a box of volume V = 1.00 m³ 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)

same as the pressure initially

2 times the pressure initially

10 times the pressure initially

20 times the pressure initially

Solution

20 times the pressure initially

According to Ideal gas equation

PV = nRT

$\frac{P_{2} V_{1}}{T_{2}} = \frac{P_{1} V_{1}}{T_{1}}$

⇒ $\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} \frac{T_{2}}{T_{1}}$

Here T₂ = 3000 K, T₁ = 300 K

Since H₂ splits into hydrogen atoms, therefore volume become half

i.e V₂ = $\frac{1}{2}$ V₁

∴ $\frac{P_{2}}{P_{1}} = \frac{V_{1}}{\frac{1}{2} V_{1}}$

∴ $\frac{P_{2}}{P_{1}}$ = 20

Some More Questions From Kinetic Theory Chapter

For Daily Free Study Material Join wiredfaculty