Thirty (30.0) milliliters of ethane (C2H6) occupy a container at STP. If the container is heated to 300.0 K and the pressure reduced to 75.0 kPa, what is the new volume?

44.6millilitres

Explanation:

Using the general gas law equation as follows:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

V1 = initial volume (L)

T1 = initial temperature (K)

P2 = final pressure (atm)

V2 = final volume (L)

T2 = final temperature (K)

According to this question;

V1 = 30mL

T1 = 273K (STP)

P1 = 1 atm (STP)

V2 = ?

T2 = 300K

P2 = 75.0 kPa = 75 × 0.00987 = 0.74atm

Using P1V1/T1 = P2V2/T2

1 × 30/273 = 0.74×V2/300

30/273 = 0.74V2/300

Cross multiply

300 × 30 = 273 × 0.74V2

9000 = 202.02V2

V2 = 9000/202.02

V2 = 44.55

V2 = 44.6millilitres.

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.