Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 0.082 atm dm3 K-1 mol-1]

A.

2.536

B.

1.623

C.

4.736

D.

0.394

Correct answer is A

For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm3

Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm3k-1 mol

Recall from ideal gas equation

pv = nRT

n = \(\frac{RV}{RT}\)

n = \(\frac{6 \times 10.5}{0.082 \times 303}\)

n= 2.536mol