Question

What volume is occupied by 0.250 mol of carbon dioxide gas at 25°C
and 0.488 atm?
A) 25.5 L B) 12.5L C) 12.5K D) 15.9L

Answer 1

Answer:

B) 12.5L

Explanation:

-We apply the mixed Gas Law:

$PV=nRT$

Where:

$P$ is the given pressure of the gas

$V$ is the volume of the gas

$n$ is the number of moles of the gas

$R, \ T$ is the ideal gas constant and temperature respectively.

$R=0.0821\ atm.L/mol.K$

#We make V the subject of the formula and substitute to solve for it;

$V=\frac{nRT}{P}\\\\T=25\textdegree C=298.15\ K\\\\\\=\frac{0.25\ mol\times 0.0821\ atm.L/mol.K\times298.15K }{0.488\ atm}\\\\\\=125.54\ L\approx 12.5\ L$

Hence, the volume of the gas is 12.5 L

Answer 2

The volume of gas occupied has been 12.5 L. Thus, option B is correct.

The carbon dioxide has been assumed to be an ideal gas. The ideal gas equation has been given as:

$PV=nRT$

Computation for the volume of Carbon dioxide

The given gas has moles of, $n=0.250\;\rm mol$

The value of pressure has been, $P=0.488\;\rm atm$

The temperature of the gas has been, $T=25^\circ \rm C\\\textit T=298\;\rm K$

The value of Boltzmann constant has been, $R=0.0821\;\rm atm.L/mol.K$

Substituting the values for the calculation of volume, (V):

$0.488\;\times\;V=0.250\;\times\;0.0821\;\times\;298\\\\V=\dfrac{0.250\;\times\;0.0821\;\times\;298}{0.488}\;\text L\\\\ V=12.53\;\rm L$

The volume of gas occupied has been 12.5 L. Thus, option B is correct.

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