Question

A sample of argon has a volume of 205 mL when its temperature is -44.0 oC and its pressure is 712 mm of Hg. What would be the volume of the argon at STP?

Answer 1

Answer:

231.98 mL

Explanation:

V₁ = Initial volume = 205 mL

V₂ = Final volume

P₁ = Initial pressure = 712 mmHg

P₂ = Final pressure = 750 mmhg (STP)

T₁ = Initial temperature = -44 °C = 229.15 K

T₂ = Final temperature = 273.15 K (STP)

From Combined gas law

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow V_2=\frac{P_1V_1T_2}{T_1P_2}\\\Rightarrow V_2=\frac{712\times 205\times 273.15}{229.15\times 750}=231.98\ mL$

Volume of the argon at STP is 231.98 mL

Answer 2

Answer:

The volume of the argon at STP is 228.95 mL.

Explanation:

Given that,

Initial volume = 205  mL

Initial temperature = -44.0 °C = 229\ K

Final temperature = 273 K

Initial pressure = 712 mm Hg

Final pressure = 760 mm Hg

We need to calculate the volume of the argon at STP

Using gas law

$\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$

Put the value into the formula

$\dfrac{712\times205}{229}=\dfrac{760\times V_{2}}{273}$

$V_{2}=\dfrac{712\times205\times273}{229\times760}$

$V_{2}=228.95\ mL$

Hence,  The volume of the argon at STP is 228.95 mL.