B. If the initial pressure of gas inside the balloon was 1.65 atm, what is the new pressure? 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?

A container of gas is heated from 250 K to 303 K. What is the new pressure if the initial pressure

is 880 kPa?

A sample of gas at 62.0°C occupies a volume of 4.80 L at a pressure of 2.90 atm. How many moles of gas are contained in the sample?

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 1 :

V1/T1 = V2/T2

3.0L/273K = V2/373K

To get the value of Z, cross multiply

3.0L x 373K = 273K x V2

1119 = 273V2

Divide both sides by 273

1119/273 = 273V2/273

4.10L = V2

The new volume is 4.10 liters

Question 2 :

P1/T1 = P2 /T2

P1 = 880 kPA= 880 *10^3 Pa

T1 = 250 K

T2 = 303 K

P2 =?

Substituting for P2

P2 = P1 T2/ T1

P2 = 880 kPa * 303 / 250

P2 = 266,640 kPa/ 250

P2 = 1066.56 kPa.

The new pressure of the gas is 1066.56 kPa

Question 3 :

Given that:

Volume of gas V = 4.80L

(since 1 liter = 1dm3

4.80L = 4.80dm3)

Temperature T = 62°C

Convert Celsius to Kelvin

(62°C + 273 = 335K)

Pressure P = 2.9 atm

Number of moles of gas N = ?

Apply ideal gas equation

pV = nRT

2.9atm x 4.8dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 335K)

13.92 atm dm3 = nx 2.747 atm dm3 mol-1

n = 13.92/2.747

n = 5.08 moles

There are 5.08 moles of gas contained in the sample

Question 4 :

Volume of gas V = 3.47L

(since 1 liter = 1dm3

3.47L = 3.47dm3)

Temperature T = 85.0°C

Convert Celsius to Kelvin

(85.0°C + 273 = 358K)

Pressure P = ?

Number of moles of gas N = 0.100 mole

Apply ideal gas equation

pV = nRT

p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)

p x 3.47dm3 = 0.29 atm dm3

p = (0.29 atm dm3 / 3.47 dm3)

p = 0.085 atm

If 1 atm = 760 mm Hg

0.085atm = 0.085 x 760

= 64.6 mm Hg

The pressure of the gas is 64.6 mm hg

answerto predict;