Calculate the volume occupied by 272g
of methane at a pressure of 250 kPa and a
temperature of 54°C.
(R = 8.31JK-1 mol-1; M, methane = 16.0)

The answer for the following problem is mentioned below.

Therefore volume occupied by methane gas is  184.78 × 10^-3 liters

Explanation:

Given:

mass of methane() = 272 grams

pressure (P) = 250 k Pa =250×10^3 Pa

temperature(t) = 54°C =54 + 273 = 327 K

Also given:

R = 8.31JK-1 mol-1 ,

Molar mass of  methane() = 16.0  grams

We know;

According to the ideal gas equation,

P × V = n × R × T

here,

n = m÷M

n =272 ÷ 16

n = 17 moles

Therefore,

250×10^3 × V = 17 × 8.31 × 327

V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )

V = 184.78 × 10^-3 liters

Therefore volume occupied by methane gas is  184.78 × 10^-3 liters

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