Question

The above shows balloon full of gas which has a volume of 120 mL at 300° kelvin swimming pressure remains constant what is the volume of the balloon if the temperature of the gas increases to 320° kelvin

Answer 1

Start off with what you are given.

V

1

: 2.75L

T

1

:

18

∘

C

V

2

: ?

T

2

:

45

∘

C

If you know your gas laws, you have to utilise a certain gas law called Charles' Law:

V

1

T

1

=

V

2

T

2

V

1

is the initial volume,

T

1

is initial temperature,

V

2

is final volume,

T

2

is final temperature.

Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.

Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:

2.75

291

=

V

2

318

By cross multiplying, we have...

291

V

2

= 874.5

Dividing both sides by 291 to isolate

V

2

, we get...

V

2

= 3.005...

In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.

Therefore,

V

2

= 3.01 L

Answer 2

124.18cm³ or 124.18mL

Step-by-step explanation:

From Charles law, the volume of a given mass of gas is inversely proportional to it Temperature provided pressure remains constant.

V1 = 120mL = 120cm³

T1 = 300°C = (300 + 273.15)K = 573.15K

T2 = 320°C = (320 + 273.15)K = 593.15K

V2 = ?

V1 / T1 = V2 / T2

V2 = (V1 * T2) / T1

V2 = (120 * 593.15) / 573.15

V2 = 124.18cm³

V2 = 124.18mL

Note 1mL = 1*10^-3L = 1cm³