Question

determine the pressure change when a constant volume of gas at 1.50 atm is heated from 20 degrees to 30 degrees

Answer 1

Answer:

This problem involves Gay-Lussac's Law. It states that the pressure exerted on the sides of a container by an ideal gas of fixed volume is proportional to its temperature.

P

1

T

1

=

P

2

T

2

P

1

= 1.00 atm;

T

1

= (20.0 + 273.15) K = 293.2 K.

P

2

= ?;

T

2

= (30.0 + 273.15) K = 303.2 K.

We know

P

1

,

T

1

, and

T

2

. Thus, we can calculate

P

2

.

P

2

=

P

1

×

T

2

T

1

= 1.00 atm ×

303.2

K

293.2

K

= 1.03 atm

Δ

P

=

P

2

–

P

1

= (1.03 -1.00) atm = 0.03 atm

The pressure increases by 0.03 atm.

This makes sense. The temperature increases by 10 parts in 300 or 3 parts in 100 (3 %). So the pressure should increase by about 3 % (0.03 atm).

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