A rigid container of O 2 has a pressure of 355 kPa at a temperature of 619 K. What is thepressure at 253 K?

130.2kPa

Explanation:

According to Gay-Lussac's Law:

P

∝

T

.

It follows that

P

=

k

T

, and so:

P

T

=

k

. It means that we can also say:

P

1

T

1

=

P

2

T

2

We need to find

P

2

. Rearranging to solve for such:

P

2

=

P

1

T

2

T

1

We know that:

P

1

=

340

kPa

T

1

=

713

K

T

2

=

273

K

Inputting:

P

2

=

340

⋅

273

713

P

2

=

130.2

kPa

is the new pressure

7.693 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

where,

are initial pressure and volume.

are final pressure and volume.

We are given:

Putting values in above equation, we get:

Thus new volume of the gas is 7.693 L