What volume does 2800mg of nitrogen gas occupy at 98Kpa and -10 C?

2.23L

Explanation:

Using the general gas law as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

Mass of nitrogen = 2800mg

Since 1g = 1000mg

2800mg = 2800/1000

= 2.8g

Using mole = mass ÷ molar mass

Molar mass of nitrogen gas (N2) = 28g/mol

mole = 2.8/28

mole of N2 gas = 0.1mol

Pressure = 98kPa

1 kilopascal (kPa) = 1000pascal (pa)

98kPa = 98 × 1000

= 98000pascal

Since 1 Pascal = 9.869 × 10^-6 atmosphere (atm)

98000pascal = 98000 × 9.869 × 10^-6

= 0.967atm

Temperature = -10°C

Kelvin = °C + 273

Kelvin = -10 + 273

K = 263K

Hence, using PV = nRT

0.967 × V = 0.1 × 0.0821 × 263

0.967V = 2.159

V = 2.159/0.967

V = 2.23

Volume of Nitrogen gas = 2.23 L

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