📖 Samacheer Kalvi · 11th TN - English Medium · Chemistry Volume 1 · Page 278question

Unit - 6 Gaseous State · Part 4

Chapter 3: 7 · Chemistry Volume 1

nRT bar dm K mol K - - bar dm Option (c) . T T = 2T V V = 2V P P = ? P V P V P P V P V P P Option (c) . γ γ H C H C H H C H M M M Squaring on both sides and rearranging × = M CnH2n- = n( ) + (2n- )( ) = 12n + 2n - = 14n - n = ( + )/ = / = Option (b) .

γ γ γ γ O H H O O H M M The fraction of oxygen that escapes in the time required for one half of the hydrogen to escape is / Option (c) . For an ideal gas PV nRT P P ∂ ∂       ∂       ∂  [ nRT]            ∂ ∂      = P nR PV nR nRT Option (b) 11th Std Chemistry 11th Std Chemistry - - - - . Greater the 'a' value, easier the liquefaction . an /V = atm a = atm L /mol = L mol - atm nb = L b = L/mol = L mol - Option (c) .

Correct Statement: Critical temperature of CO is K. It means that CO cannot be liquefied above K, whatever the pressure may applied. Pressure is inversely proportional to volume Option (d) . Density = = Molar mass × Mass m Volume nRT ⎛ ⎝ ⎛ ⎝ P m ⎛ ⎝ ⎛ ⎝ P RT P RT g mol - × atm .

L atm K - mol - × K = .41g L - Option (c) . For a fixed mass of an ideal gas V α T P α

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