Question

A sample of argon gas has a volume of 735 mL at a pressure of 1.20 atm and a temperature of 112 ∘C. What is the final volume of the gas, in milliliters, when the pressure and temperature of the gas sample are changed to the following, if the amount of gas does not change?Part A) 660 mmHg and 274 K

Answer 1

A) 722.6 mL

Procedure

To solve this problem we can use the ideal gas Law given the temperature and pressure conditions.

Data

R= 0.08206L⋅atm⋅K⁻¹⋅mol⁻¹

P= 1.20 atm

T=112 °C =385.15 °K

V=0.735 L

Formula

PV=nRT

Solving for n to get the moles we will have

`n=(PV)/(RT)=\frac{1.20\text{ atm }0.735\text{ L }\degree\text{K.mol}}{0.08206\text{ L.atm }385.15\text{ }\degree\text{K}}=0.02791\text{ mol}`

Once we have calculated the moles we solve again with the new temperature and pressure conditions, remember to use the units of R constant. Pressure in atm and temperature in Kelvin for this case.

Data

n=0.02791 mol

R= 0.08206L⋅atm⋅K⁻¹⋅mol⁻¹

P= 660 mmHg = 0.868421 atm

T=274 °K

Solve for V

`V=(nRT)/(P)=\frac{0.02791\text{ mol }0.08206\text{ L.atm }274\degree\text{K}}{0.868421\text{ atm mol }\degree\text{K}}=0.7226\text{ L}=722.6\text{ mL}`