Question

1. The pressure of 5.85 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is decreased by one-half. What is the final volume of the gas?

2. A piece of sodium metal reacts completely with water as follows:

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

The hydrogen gas generated is collected over water at 27.0°C. The volume of the gas is 175 mL measured at 0.969 atm. Calculate the number of grams of sodium used in the reaction. (The vapor pressure of water at 27.0°C = 0.0352 atm.)

Answer 1

1. we know the equation

P₁V₁/T₁ = P₂V₂/T₂ where,

P₁ = initial pressure = consider x

V₁ = initial volume = 5.85 L

T₁ = initial tempreture = consider y

P₂ = final presssure = x/3

V₂ = final volume = ?

T₂ = final tempreture = y/2

we can write above equation

V₂ = P₁V₁T₂/P₂T₁  substiture value in this equation

V₂ = (x) (5.85) (y/2) / (x/3) (y) same value get cancelled then

V₂ = 5.85 3 /2 = 8.775 L

final volume of gas is 8.775 L

2.

Use ideal gas equation for calculation of mole of hydrogen gas

Ideal gas equation

PV = nRT             where, P = atm pressure= 0.969 atm,

V = volume in Liter = 175 ml = 0.175 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 27⁰ = 273.15+ 27 = 300.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.9690.175)/(0.08205300.15) = 0.0068856 mole of H₂

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

according to this reaction to prepair 1 mole of H₂ require 2 mole of Na

then to prepare 0.0068856 mole of H₂ require 0.0068856 2 = 0.01377 mole of Na

molar mass of Na = 22.9897gm/mol

1 mole of Na = 22.9897 gm then 0.01377 mole of Na = 22.9897 0.01377 = 0.3165 gm

0.3165 gm of sodium metal used in the reaction.