Calculate the density of krypton gas at a pressure of 755 mmHg and a temperature of 38 ° C.

Answer 1

To calculate the density of krypton gas at a given pressure and temperature, we can use the ideal gas law equation:

PV = nRT

where:
P = pressure (in atm)
V = volume (which can be assumed to be 1 L for simplicity)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)

First, we need to convert the given pressure of 755 mmHg to atm:

1 atm = 760 mmHg

So, 755 mmHg is equal to 0.99342 atm.

Next, we convert the given temperature of 38 °C to Kelvin:

T(K) = T(°C) + 273
T = 38 + 273
T = 311 K

Now, we can rearrange the ideal gas law equation to solve for density (d):

n/V = P/RT

Since V = 1 L, n/V represents the density (d) of the gas.

Substitute the values into the equation:

d = (0.99342 atm) / (0.0821 L.atm/mol.K * 311 K)
d ≈ 0.0320 mol/L

The molar mass of krypton (Kr) is approximately 83.80 g/mol.

Now we can convert the molar density to mass density using the molar mass:

molar density = 0.0320 mol/L
mass density = 0.0320 mol/L * 83.80 g/mol
mass density ≈ 2.68 g/L

Therefore, the density of krypton gas at 755 mmHg and 38 °C is approximately 2.68 g/L.