COMEDK 2024 Shift 1 Paper

To determine the total mass of the gas in the box, we first need to understand the given problem and apply the principles of kinetic theory of gases. Let's summarize the given data:
Side length of the cubical box, L=1m
Pressure of the gas, P=100Nm−2
Observation time, t=1 s
Number of hits on the wall, n=500
We are given that the atom travels with the root mean square (rms) speed (denoted by vrms ) parallel to one edge of the cube, and makes 500 hits on the wall in 1 second. From this, we can find the vrms as follows:
Each hit corresponds to the atom travelling across the side of the cube and bouncing back. The distance travelled between two consecutive hits is 2L, as the atom goes to the wall and returns. Therefore, the total distance travelled in 1 second is:

500×2L=500×2×1=1000m
So, the rms speed vrms is:
vrms=‌

1000m

1 s

=1000m∕ s
Now, using the formula from the kinetic theory of gases for pressure:
P=‌

1

3

ρvrms2
where ρ is the density of the gas. We can express density ρ as:
ρ=‌

m

V

Here, m is the mass of the gas and V is the volume of the box. For a cubical box, V=L3, so:
V=13m3=1m3
Substituting the values into the pressure equation:
100Nm−2=‌

1

3

(‌

m

1

)(1000)2
Simplifying, we get:
‌100=‌

1

3

m×1000000
‌100=‌

1000000

3

m
‌m=‌

100×3

1000000

‌m=0.0003‌kg
Converting this mass to grams:
m=0.0003‌kg×1000g∕kg=0.3g
Therefore, the total mass of the gas in the box is:
Option B: 0.3 grams