At 25◦C a 4 L sample of H2 exerts a pressure of 5 atm. What pressure would the same sample exert in a 2 L container at 25◦C?

10 atm.

Explanation:

Using the combined gas law equation as follows;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 5 atm

P2 = ?

V1 = 4L

V2 = 2L

T1 = 25°C = 25 + 273 = 298K

T2 = 25°C = 298K

Using P1V1/T1 = P2V2/T2

5 × 4/298 = P2 × 2/298

20/298 = 2P2/298

Cross multiply

298 × 20 = 298 × 2P2

5960 = 596P2

P2 = 5960 ÷ 596

P2 = 10 atm.

thats how its supposed to be