What is the density of helium gas at 17 degrees celcius and 773 torr?

0.172g/L

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

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