please help im literally suffering
3. If the initial pressure of gas inside the balloon was 1.65 atm, what is the new pressure?
4. A weather balloon has a volume of 200.0 L at a pressure of 760 mm Hg. As it rises, the pressure decreases to 282 mm Hg.
a. Will the volume of the balloon increase or decrease? Why?
b. What is the new volume of the balloon? (Assume constant temperature).
5. A sample of gas kept at a constant pressure, is heated from 273 K to 373 K. If the initial volume is 3.00 L, what is the new volume?
6. A container of gas is heated from 250 K to 303 K. What is the new pressure if the initial pressure is 880 kPa? (Assume constant volume)
A 0.100 mole sample of gas is at a temperature of 85.0°C and a volume of 3.47 L. What is the pressure of the gas (in mm Hg)

Question

11-07-2022
Chemistry

contestada

please help im literally suffering
3. If the initial pressure of gas inside the balloon was 1.65 atm, what is the new pressure?
4. A weather balloon has a volume of 200.0 L at a pressure of 760 mm Hg. As it rises, the pressure decreases to 282 mm Hg.
a. Will the volume of the balloon increase or decrease? Why?
b. What is the new volume of the balloon? (Assume constant temperature).
5. A sample of gas kept at a constant pressure, is heated from 273 K to 373 K. If the initial volume is 3.00 L, what is the new volume?
6. A container of gas is heated from 250 K to 303 K. What is the new pressure if the initial pressure is 880 kPa? (Assume constant volume)
A 0.100 mole sample of gas is at a temperature of 85.0°C and a volume of 3.47 L. What is the pressure of the gas (in mm Hg)

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2022-07-12T21:12:22+02:00

The answers are as follows;

3 - 4.95 atm

4 - 539 L

5 - 4 L

What are the gas laws?

The gas laws are used to establish the relationship between the pressure, volume, number of moles etc of a gas.

3.

V1 = 1.50 L

V2 = 0.50 L

P1 = 1.65 atm

P2 = ?

P1V1 = P2V2

P2 = P1V1 /V2

P2 = (1.50)(1.65) / (0.50)

P2 = 4.95 atm

4.

P1V1 = P2V2

P1 = 760 mm Hg

V1 = 200.0 L

P2 = 282 mm Hg

V2 = ?

V2 = P1V1/P2

V2 = 760 mm Hg * 200.0 L/282 mm Hg

V2 = 539 L

The volume of the gas increases because the pressure decreases.

5.

Using

V1/T1 = V2/T2

V1T2 =V2T1

V2 = V1T2/T1

V2 = 3.00 L * 373 K/ 273 K

V2 = 4 L

Learn more about gas laws:https://brainly.com/question/22027561

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