A box of 1.00m³ is filled with nitrogen at 1.50 atm at 300K. The box has a hole of an area 0.010 mm². How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.

This is a long answer type question as classified in NCERT Exemplar

Given volume V = 1m³

area = 0.01mm²

= 8.01 $\times {10}^{-6}$ m²= ${10}^{-8}$ m²

Temperature both inside and outside are equal

So initial pressure inside the box = 1.50atm

Final pressure inside the box= 0.1atm

Assuming V_x= speed of nitrogen molecule in x direction

n_i = number of molecules per unit volume in a time interval of $?T$

Let area of the wall, number of particles colliding in time

$?t$ = $\frac{1}{2}n$ _i (v_{x $?t$} )A , here we use ½ because particle moves both in positive and negative direction.

V_x²+ V_y²+ V_z²= V_rms²

V_x²= V_rms²/3 if all three velocities are equal.

½ mv_rms²= 3/2K_BT/m

<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>Given volume V = 1m³</p><p>area = 0.01mm²</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745555770phpiBUASa_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745555770phpiBUASa.jpeg" alt="" width="145" height="112"></picture></div></div><p>= 8.01<span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>6</mn> </mrow> </mrow> </msup> </math> </span>m²= <span contenteditable="false"> <math> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>8</mn> </mrow> </mrow> </msup> </math> </span>m²</p><p>Temperature both inside and outside are equal</p><p>So initial pressure inside the box = 1.50atm</p><p>Final pressure inside the box= 0.1atm</p><p>Assuming V_x= speed of nitrogen molecule in x direction</p><p>n_i = number of molecules per unit volume in a time interval of <span contenteditable="false"> <math> <mo>? </mo> <mi>T</mi> </math> </span></p><p>Let area of the wall, number of particles colliding in time</p><p><span contenteditable="false"> <math> <mo>? </mo> <mi>t</mi> </math> </span>=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>n</mi> </math> </span>_i (v_{x<span contenteditable="false"> <math> <mo>? </mo> <mi>t</mi> </math> </span>})A&nbsp;, here we use ½ because particle moves both in positive and negative direction.</p><p>V_x²+ V_y²+ V_z²= V_rms²</p><p>V_x²= V_rms²/3 if all three velocities are equal.</p><p>½ mv_rms²= 3/2K_BT/m</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745558193phpPzfxU4_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745558207phpHaKk8k_480x360.jpeg" alt="" width="434" height="315"></picture></div></div><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745558207phpHaKk8k_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745558207phpHaKk8k.jpeg" alt="" width="269" height="305"></picture></div></div>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment