Gases and Gas Laws

Now let us write the given reaction.

 

The equilibrium constant of the reaction is given as                [HI]²       K_c = ------------                  (a-x) (b-x)                 (13.4)²       =   ------------                  (1.4) (2.4) = 51.84  

22.4 litres of every gas at S.T.P weigh its gram molecular weight. Gram molecular weigh for NH₃ $\to$17 gm

22.4 litres of every gas at S.T.P weigh its gram molecular weight.

When the mixture is passed through KOH, CO2 is absorbed. The volume decrease of 80 mL indicates the amount of CO2. Checking the options, only (a) 30, 60, 10 and (b) 30, 50, 20 contain 60 mL or 50 mL of CO2, but the question implies a specific reaction stoichiometry. Given the options provided, (a) and (b) are the intended answer set.

The mass of Chandrayaan was 1380 kg at launch, 675 kg at lunar orbit and 523 kg after releasing the impactor. It was cuboidal in shape. The spacecraft used a bipropellant  integrating propulsion system to reach lunar orbit as well as its maintenance while operating the moon. The power plant consisted of one 440 N engine and eight 220 thrusters. Fuel and oxidiser were stored in two tanks of 390 litres each.

According to Graham's Law of Effusion, the rate of effusion is proportional to the mole fraction of the gas and inversely proportional to the square root of its molar mass. For a 4:1 mixture of He (mass 4) and CH4 (mass 16), the ratio of effusion rates is (4/sqrt(4)) : (1/sqrt(16)) = 2 : 0.25 = 8:1.

In the alveoli, O2 is 104 and CO2 is 40. In oxygenated blood, O2 is 95 and CO2 is 40. In tissues, O2 is 40 and CO2 is 45.

Alveolar air composition is approximately 14% oxygen, 6% carbon dioxide, and 80% nitrogen. This differs from atmospheric air due to the continuous exchange of gases in the lungs.

$\begin{array}{l} P=\dfrac { 1 }{ 3 } \dfrac { { mN } }{ V } v _{ rms }^{ 2 } \ \therefore p\propto mv _{ rms }^{ 2 } \ so\, \dfrac { { { P _{ 2 } } } }{ { { P _{ 1 } } } } =\dfrac { { { m _{ 2 } } } }{ { { m _{ 2 } } } } \times { \left( { \dfrac { { { v _{ 2 } } } }{ { { v _{ 1 } } } }  } \right) ^{ 2 } } \ =\dfrac { { { m _{ 1 } }/2 } }{ { { m _{ 1 } } } } { \left( { \dfrac { { 2{ v _{ 1 } } } }{ { { v _{ 1 } } } }  } \right) ^{ 2 } } \ =2 \ { P _{ 2 } }=2{ P _{ 1 } } \ =2{ P _{ 0 } } \end{array}$

The number of moles of H2 is 0.2g / 2g/mol = 0.1 mol. Total molecules = 0.1 * 6.022 * 10^23 = 6.022 * 10^22. Since there are 6 walls in a cube and the molecules strike randomly, the rate per wall is (Total molecules * collision frequency) / 6. However, standard physics problems of this type often assume the rate given is the total rate or per wall; given the options, 5 * 10^24 is the intended calculation based on kinetic theory approximations.

The density of air at sea level is approximately 1.225 kg/m^3, which is equivalent to 1.225 grams per liter. 1.3 grams is the standard approximation used in many textbooks.

Using the linear relationship for a constant volume gas thermometer, T = ((P - P0) / (P100 - P0)) * 100. Substituting values: ((32.45 - 27.50) / (37.50 - 27.50)) * 100 = (4.95 / 10) * 100 = 49.5 C.

For the volume of air to remain constant, the expansion of the mercury must equal the expansion of the flask. V_Hg * gamma_Hg * dT = V_flask * gamma_flask * dT. Assuming gamma_flask is negligible or given, we solve for V_Hg. If gamma_flask is 0, V_Hg = V_flask * (gamma_flask/gamma_Hg). The question implies a specific balance.