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1.0 moles of an ideal gas are held in a container under 2.0 atm of pressure and 310 K. What must be the volume of this container?

Respuesta :

Answer:

12.7 L (3 s.f.)

Explanation:

Since it is an ideal gas, we can apply the ideal gas law:

[tex]\boxed{pV=nRT}[/tex]

☆ where p= pressure

V= volume

n= number of moles

R= ideal gas constant

T= temperature

Substitute all the given information into the formula:

2V= 1(0.08206)(310)

2V= 25.4386

V= 25.4386 ÷2

V= 12.7 L (3 s.f.)

Further explanation:

The ideal gas constant, R, has different values depending on what units are being used. Two examples are listed below:

[tex]• \: 0.08206 \: L \: atm ^{ - 1} \: mol \: K[/tex]

[tex]• \: 8.314 \: J \: {mol}^{ - 1} \: {K}^{ - 1} [/tex]

In the above question, we use the value 0.08206 because the pressure was given in atm and the temperature was given in Kelvin. Thus, the unit used for volume is L.

Answer: 12.7 L

Explanation:

P=2.0atm

V= ???

n=1.0 mol

R= 0.0821

T= 310K

PV=nRT--> (2)V=(1)(0.0821)(310)

                     V=12.7 L