Question

0.10 mol of argon gas is admitted to an evacuated 50 cm3 container at 20°C. The gas then undergoes heating at constant volume to a temperature of 300°C. What is the final pressure of the gas? M - 106

Answer 1

Answer: The final pressure of the gas is 9.41 atm

Explanation:

To calculate the pressure of the gas, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = pressure of the gas = ?

V = Volume of gas = $50cm^3=0.05L$    (Conversion factor:  $1L=1000cm^3$ )

n = Number of moles = 0.01 mol

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

T = temperature of the gas = $300^oC=[300+273]K=573K$

Putting values in above equation, we get:

$P\times 0.050L=0.01\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 573K\\\\P=\frac{0.01\times 0.0821\times 573}{0.05}=9.41atm$

Hence, the final pressure of the gas is 9.41 atm

Answer 2

The final pressure of the gas is 94.09atm

According to the general gas equation;

$PV = nRT$

• P is the pressure of the gas

• V is the volume of the gas = 50cm³ = 0.05L

• n is the number of moles of element = 0.10mol

• R is the gas constant = 0.0821 L atm/molK

• T is the temperature = 300°C. + 273 = 573K

Substitute the given parameters into the formula:

$P=\frac{nRT}{V} \\P=\frac{0.1\times 0.0821 \times 573}{0.05}\\P = \frac{4.70433}{0.05} \\P=94.09atm$

Hence the final pressure of the gas is 94.09atm

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