ASAP Step by step I really am not understanding how to solve this problem

Question

If 10 g of oxygen gas is in a 2.00 L container at 50 Cwhat is the pressure in the container ? Round to one decimal place.

When the temperature of a rigid hollow sphere containing 685 L of helium gas is held to 62C, the pressure of the gas is 1.89 * 10 ^ 3 * kPa . How many grams of helium does the sphere contain? (Round to 3 significant digits)

Answer 1

Do you know the ideal gas law? If not, you must learn it.

PV = nRT

P = pressure = ?

V = volume in liters = 2.00 L

n = moles of O2 = 10 g x 1 mol O2 / 32 g = 0.3125 mols O2

R = gas constant = 0.0821 Latm/Kmole

T = temperature in Kelvin = 50C + 273 = 323K

Solve for P

P = nRT / V = (0.3125)(0.0821)(323) / 2.00

P = 4.14 atm = 4.1 atm (to 1 decimal place)

PV = nRT

solve for n (moles of helium)

n = PV / RT

n = ?

P = pressure = 1.89x10³ kPa ...converted to atm we get 1.89x10³ kPa x 1 atm / 101.325 kPa = 18.65 atm

V = volume = 685 L

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 62C + 273 = 335K

n = PV/RT = (18.65)(685) / (0.0821)(335)

n = 465 moles (3 sig. figs.)

Convert this to grams using molar mass of He (4 g/ mol)

mass of helium = 465 mols x 4 g/mol = 1860 g (3 sig. figs.)

Be sure to check the math