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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :