Part 1. determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. show your work. part 2. if this sample was placed under extreme pressure, describe how the actual volume would compare to the predicted volume. explain your answer.

Part 1: 9.165 g/mol.

Part 2: The volume will decrease.

Explanation:

Part 1:

PV = nRT,

where, P is the pressure of the gas (atm) (P = 0.98 atm).

V is the volume of the gas (L) (V = 1.2 L).

n is the no. of moles of the gas (mol).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature of the gas (K) (T = 287 K).

∴ n = PV/RT = (0.98 atm)(1.2 L)/(0.082 L.atm/mol.K)(287 K) = 0.0499 mol ≅ 0.05 mol.

∵ n = mass/molar mass

∴ molar mass = mass/n = (0.458 g)/(0.05 mol) = 9.165 g/mol.

Part 2:  

The volume of the gas will decrease as the sample is placed under extreme pressure.

Boyle’s Law  states that at constant temperature, the  volume of a given quantity of  a gas varies inversely with its  pressure.

P ∝ 1/V.

so, as the sample is placed under extreme pressure, the volume of the gas will be decreased.

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

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.