Normal 0false false falseEN-US X-NONE X-NONE/* Style Definitions */table.MsoNormalTable{mso-style-name:Table Normal;mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-parent:;mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin-top:0in;mso-para-margin-right:0in;mso-para-margin-bottom:10.0pt;mso-para-margin-left:0in;line-height:115%;mso-pagination:widow-orphan;font-size:11.0pt;font-family:Calibri,sans-serif;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:Times New Roman;mso-bidi-theme-font:minor-bidi;}N molecules each of mass m of a gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel which is maintained at temperature T. The mean square velocity of molecules of B type is v2 and the mean square rectangular component of the velocity of A type is denoted by w2. Then w2/v2(a) 2(b) 1(c) 1/3(d) 2/3
Hrishant Goswami , 11 Years ago
Grade 10
1 Answers
Amit Saxena
Last Activity: 11 Years ago
(d)
Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules = w2 + w2 + w2 = 3w2.
Therefore,
This gives m2/v2 = 2/3
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