What is the volume of 10.0 g of argon gas at 157◦C and 2.50 kPa pressure?

Answer 1

This is solved using the ideal gas law equation PV = nRT

I like to make a little table of given amounts and the constant (R) to make sure I am consistent with units

R = 8.314 x 10³ L x Pa K^-1mol^-1 or it can be written 8.314 x 10³ L x Pa

K x mol

P = 2.5 kPa (but I need pressure in Pa) 2.5kPa x 1000 Pa/1 kPa = 25000 Pa

V = ?

n = moles? (but I'm given grams) 10 g Ar x 1 mol Ar /39.9 g Ar = 0.2506 mol Ar (using Molar Mass of Ar)

T = 157 C (but I need temperature in K) 157C +273 = 430 K

Plugging these into PV = nRT or V = nRT/P will give you your final answer. Pay attention to sig figs as needed.