Question

What volume will 1.27 moles of helium gas occupy at 80.00 °C and 1.00 atm?

Answer 1

Answer:

36.8 L

Explanation:

We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 80 °C

T(K) = 80 + 273

T(K) = 353 K

Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:

Number of mole (n) = 1.27 moles

Temperature (T) = 353 K

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

PV = nRT

1 × V = 1.27 × 0.0821 × 353

V = 36.8 L

Thus, the volume occupied by the helium gas is 36.8 L