Question

What will be the volume of one mole of oxygen at 27°C
and at a pressure twice the standard atmosphere.
​

Answer 1

Explanation:

The volume will be

6.15

liters.

Explanation:

We use the ideal gas equation, which states that

P

V

=

n

R

T

We need to solve for volume, so rearrange the equation into

V

=

n

R

T

P

We use the given values for this question, which is

n

=

1

mol

R

=

0.082

L

atm

K

−

1

mol

−

1

T

=

300

K

P

=

4

atm

We get:

V

=

1

mol

⋅

0.082

L

atm

K

−

1

mol

−

1

⋅

300

K

4

atm

=

6.15

L

So, the volume of the oxygen gas will be

6.15

liters or

6150

milliliters.

Of course, this would not be the exact value, as no gas behaves exactly like an ideal gas.