Question

calculate the volume occupied by 3.5g of oxygen fast at 27°c and 740mm pressure

Answer 1

According to PV=nRT
Here P=740 mm = 740/760 bar = 0.97368 bar
n of O2 is 3.5/32 = 0.109375 mole
R=0.08314
T= 27°C = 300 k
Hence V= .109375 * 0.08314 * 300 /0.97368
V= 2.80 litre

Answer 2

Answer: The volume occupied by oxygen gas is 2.757L

Explanation:

To calculate the volume occupied by given amount of gas, we use the ideal gas equation, which is:

$PV=nRT$

where.

P = pressure of the gas = $\frac{740mmHg}{760mmHg}\times 1atm=0.9736atm$    (Conversion Factor: 1 atm = 760mmHg)

V = volume occupied by this gas = ?L

T = temperature of the gas = [27 + 273]K = 300K

n = Number of moles of oxygen gas = $\frac{3.5g}{32g/mol}=0.109mol$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

Putting values in above equation, we get:

$0.9736atm\times V=0.109mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\V=2.757L$

Hence, the volume occupied by oxygen gas is 2.757L