Two flasks I and II shown below are connected by a value of negligible volume. 2.8gM2 300 K 1 L I II 0.2g N2 60 K 2 L When the value, the final pressure of the system in bar is x × 10–2. The value of

Question

Two flasks I and II shown below are connected by a value of negligible volume. 2.8gM2 300 K 1 L I II 0.2g N2 60 K 2 L When the value, the final pressure of the system in bar is x × 10–2. The value of x is _______. (Integer answer) [Assume : Ideal gas; 1 bar = 105 Pa : Molar mass of N2 = 28.0 mol–1; R = 8.31 J mol–1 K–1] ¶ykLd I rFkk II ,d ux.; vk;ru ds okYo ls tqMs+ gSa % 2.8gM2 300 K 1 L I II 0.2g N2 60 K 2 L okYo dks [kksy nsus ij fudk; esa vafre nkc (bar esa) x × 10–2 gks tkrk gSA x dk eku gS _____. (fudVre iw.kk±d esa) [eku yhft, xSl vkn'kZ gS : 1 bar = 105 Pa : eksyj lagfr N2 = 28.0 mol–1; R = 8.31 J mol–1 K–1]

Prashant Kotian · Accepted Answer

Correct answer: .. Applying ; (n1 + nII)initial = (n1 + nII)final  Assuming the system attains a final temperature of T (such that 300 < T < 60)  {|Heat lost by N2 of container (I)| = |Heat gained by N2 of container II|}  n1Cm(300 – T = nIICm(T – 60)        8 . (30 – T) = 28 . (T – 60)  14(300 – T) = T – 60  ) 300 (   = T  T = 284 K (final temperature)  If the final pressure = P  (n1 + nII)final       0 . | JEE MAIN-2021 | DATE : 27-08-2021 (SHIFT-2) | PAPER-1 | OFFICAL PAPER | CHEMISTRY PAGE # 9  RT P (VI + VII) = mol / gm gm . P = Pa bar m 3 K K mol J . mol 3 3        0.84287 bar  84.28 × 10–2 bar  84

💡 Solution & Explanation

Practice 22,000+ questions like this