First slide

Kinetic theory of ideal gases

Question

1 mole of H₂ gas is contained in a box of volume V=1.00 m³ at T₁ = 300 K. The gas is heated to a temperature of T₂ = 3000 K and the gas gets converted into a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

Moderate

A

same as the pressure initially
B

2 times the pressure initially
C

10 times the pressure initially
D

20 times the pressure initially

Solution

When the molecules break into atoms, the number of moles would become twice.
Now, by ideal gas equation
P = Pressure of gas,
n = Number of moles
R Gas constant,
T Temperature
We have pV= nRT
As volume (V) of the container is constant.
As gases break, number of moles becomes twice of initial number.
so $n_{2} = 2 n_{1}$
So $p \propto n T$
$\begin{array}{l} \Rightarrow \frac{p_{2}}{p_{1}} = \frac{n_{2} T_{2}}{n_{1} T_{1}} = \frac{(2 n_{1}) (3000)}{n_{1} (300)} = 20 \\ \Rightarrow p_{2} = 20 p_{1} \end{array}$
Hence, final pressure of the gas would be 20 times the pressure initially.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App