What volume of neon gas occupies a container if 322 grams of neon is at 0.955 atm and -28.0 degrees Celsius?

The volume occupies of neon gas is 335, 7 L

Explanation:

We use the formula PV=nRT. We convert the unit temperature Celsius into Kelvin: 0°C=273 K---> -28°C= -28+273= 245K. We calculate the number of mols in 322 grams of neon:

20,18 g---1 mol neon

322g---x= (322g x 1 mol neon)/ 20, 18g=15, 96 mol neon

PV=nRT    > V= (nRT)/P

V= (15,96 mol x 0,082 l atm/K mol x 245 K)/ 0,955 atm= 335, 7 L

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm