Question

4 অ্যাটমােস্ফিয়ার চাপে ও 300K উন্নতায় 8g H,
গ্যাসের (H = 1) আয়তন কত
(R = 0.082 L• atm • mol-1. K-1)​

Answer 1

We are going to use the ideal gas law, which states that

PV=nRT

P is pressure in pascals (Pa)

V is volume in m3

n is the number of moles of the substance

R is the gas constant

T is the temperature in Kelvin (K)

Since we are using atmospheric pressure as units and the gas constant as

0.082

Lit−atm−mol−1K−1

, then our volume will be in liters (l)

So, plugging in P=4atm ,

n = 8g ÷ 2.016g/mol ≈ 3.97mol ,

R = 0.082

and

T = 300 K

we get

V = (n R T)/P =( 3.97 × 0.082×300)/4 ≈ 24.42 l = 24420 cm3 of hydrogen gas (H2)

.