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.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

answerlinnaeus;