Question

Need help with some chem equations

1. While diving under 3.7 atm of pressure, a gas bubble of 25 mL was found when release at surface at 1 at how big did the bubble get

2. a blue 3 L balloon at 25c was moved to a temperature of 38c what is new volume

3. 1.72 L of gas at STP was transferred to a new container with temps 18c and volume of 3.1 L what is new pressure

Answer 1

1. 92.5 ml

2. 3.13 L

3. 524.48 K

Step-by-step explanation:

1. Using Boyle's law equation:

P1V1 = P2V2

Where;

P1 = initial pressure (atm)

V1 = initial volume (ml)

V2 = final volume (ml)

P2 = final pressure (atm)

Based on the information in question 1: P1 = 3.7atm, P2 = 1atm, V1 = 25 mL, V2 = ?

3.7 × 25 = 1 × V2

92.5 = V2

V2 = 92.5ml

2. Using Charles law equation as follows:

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

Based on the information provided: V1 = 3L, V2 = ?, T1 = 25°C = 25 +273 = 298K, T2 = 38°C = 38 + 273 = 311K

3/298 = V2/311

Cross multiply

298 × V2 = 3 × 311

298V2 = 933

V2 = 933 ÷ 298

V2 = 3.13L

3. Using Charles law equation as follows:

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

Based on the information provided; V1 = 1.72 L, T1 = 18°C = 18 + 273 = 291K, T2 = ?, V2 = 3.1L

1.72/291 = 3.1/T2

Cross multiply

1.72 × T2 = 291 × 3.1

1.72T2 = 902.1

T2 = 902.1 ÷ 1.72

T2 = 524.48K