Option E density
Explanation:
This is actually pretty easy to explain. At the beggining it states that at STP which are 1 atm and 273 K, we have two samples of H2 but in different quantities. So using the ideal gas equation, we calculate the volume of each gas:
PV = nRT (1)
Where:
P: pressure in atm
V: volume in L
n: moles
R: gas constant which is 0.082 L atm / K mol
T: temperature in K
So from this equation, we solve for V:
V = nRT/P
Replacing data for both samples we have:
V1 = 0.5 * 0.082 * 273 / 1 = 11.19 L
V2 = 1 * 0.082 * 273 / 1 = 22.38 L
Now, to verify that is option E, let's write the expression for density:
d = m/V (2)
Where:
d: density
m: mass
To calculate the mass, we use the molar weight of hydrogen (2 g/mol) and the moles of the samples so:
m1 = 0.5 * 2 = 1 g
m2 = 1 * 2 = 2 g
Now, replacing in (2):
d1 = 1 / 11.19 = 0.0893 g/L
d2 = 2 / 22.38 = 0.0893 g/L
As d1 = d2 we can conclude that option E is the correct option.
The correct answer is option A.
Explanation:
The average kinetic energy of the gas particle only depends upon the temperature of the gas.
The formula for average kinetic energy is:
where,
k = Boltzmann’s constant =
T = temperature
So, at standard temperature and pressure 0.50 moles of hydrogen and 1.0 mole of oxygen sample will have same value of average kinetic energy.
Where as in other option enlisted in question , molar masses of both gases will be involved which will give different answers for both the gases.
The correct answer is option A.
Explanation:
The average kinetic energy of the gas particle only depends upon the temperature of the gas.
The formula for average kinetic energy is:
where,
k = Boltzmann’s constant =
T = temperature
So, at standard temperature and pressure 0.50 moles of hydrogen and 1.0 mole of oxygen sample will have same value of average kinetic energy.
Where as in other option enlisted in question , molar masses of both gases will be involved which will give different answers for both the gases.