Answer:
0.172g/L
Step-by-step explanation:
Step 1:
Data obtained from the question:
Temperature (T) = 17°C
Pressure (P) = 773 torr
Step 2:
Conversion to appropriate unit:
For pressure :
760 torr = 1 atm
Therefore, 773 torr = 773/760 = 1.02 atm
For temperature:
Temperature (Kelvin) = temperature (celsius) + 273
temperature (celsius) = 17°C
Temperature (Kelvin) = 17°C + 273 = 290K.
Step 3:
Obtaining an expression for the density.
From the ideal gas equation PV = nRT, we can obtain an equation for the density as follow:
PV = nRT. (1)
But: number of mole(n) = mass (m)/Molar Mass(M) i.e
n = m/M
Substitute the value of n into equation 1
PV = nRT
PV = mRT/M
Divide both side by m
PV /m = RT/M
Divide both side by P
V/m = RT/MP
Invert the above equation
m/V = MP/RT (2)
Recall:
density (d) = mass(m) / volume(V) i.e
d = m/V
Replace m/V in equation 2 with d
m/V = MP/RT
d = MP/RT.
Step 4:
Determination of the density.
Temperature (T) = 290K
Pressure (P) = 1.02 atm
Molar Mass of helium (M) = 4g/mol
Gas constant (R) = 0.082atm.L/Kmol
Density (d) =?
d = MP/RT
d = 4 x 1.02 / 0.082 x 290
d = 0.172g/L
Therefore, the density of the helium gas is 0.172g/L