On a recent trip to the Fort Worth Zoo, Trevor's parents bought him a helium-filled balloon. At the beginning of the day, the sky was sunny and clear. The balloon was filled with 14.2 L of helium at a pressure of 102.5 kPa. The temperature was 33 °C. After a while, a storm came up, and a drop in barometric pressure and temperature reduced the pressure of the helium to 100.9 kPa and the temperature to 24 °C. What was the new volume of the balloon if there was a change? Did the volume increase or decrease?

1. The new volume is 14L

2. The volume decreased

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 14.2 L

Initial pressure (P1) = 102.5 kPa

Initial temperature (T1) = 33°C

Final pressure (P2) = 100.9 kPa

Final temperature (T2) = 24°C.

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

K = °C + 273

T1 = 33°C = 33°C + 273 = 306K

T2 = 24°C = 24°C + 273 = 297K

Step 3:

Determination of the new volume.

Applying the general gas equation P1V1/T1 = P2V2/T2, the new volume can be obtain as follow:

P1V1/T1 = P2V2/T2

102.5 x 14.2/306 = 100.9 x V2/297

Cross multiply to express in linear form as shown below:

306 x 100.9 x V2 = 102.5 x14.2 x297

Divide both side by 306 x 100.9

V2 = (102.5x14.2x297)/(306 x 100.9)

V2 = 14 L

The new volume is 14 L.

Step 4:

Determination of the change in volume. This is illustrated below

Final volume (V2) = 14 L

Initial volume (V1) = 14.2 L

Change in volume (ΔV) =?

Change in volume (ΔV) = Final volume (V2) - Initial volume (V1)

ΔV = V2 - V1

ΔV = 14 - 14.2

ΔV = - 0.2L

Since the change in volume is negative, it means there is a decrease in the volume.

The volume decreased to 14.0 L

Explanation:

Step 1: Data given

Volume of the balloon 14.2 L

Pressure = 102.5 kPa = 1.0115963 atm

Temperature = 33 °C = 306 K

The pressure reduced to 100.9 kPa = 0.99580558

Temperature reduced = 24.0 °C = 297 K

Step 2: Calculate the new volume

(P1*V1)/T1 = (P2*V2)/T2

⇒with P1 is the initial pressure = 102.5 kPa

⇒with V1 = the initial volume of the balloon = 14.2 L

⇒with T1 = the initial temperature = 306 K

⇒with P2 = the reduced pressure = 100.9 kPa

⇒with V2 = the final volume = TO BE DETERMINED

⇒with T2 = the reduced temperature = 297 K

(102.5 kPa * 14.2 L) . 306 K  = (100.9 kPa * V2) / 297 K

4.7565 = (100.9 *V2) / 297 K

1412.68 = 100.9 * V2

V2 = 14.00 L

The volume decreased to 14.0 L

out of these options:

ability to burn paper

pure sodium reacting violently with water

the boiling point of water

helium does not tend to react with anything

i would say this is a physical property:

helium does not tend to react with anything

why? :

i said helium does not tend to react with anything because well, it reacts to ! but ability to burn paper, pure sodium reacting violently with water, & the boiling point of water are chemical or

physical properties: the properties of a substance that can be observed without changing the chemical makeup of the substance

chemical properties: properties of a substance relating to the chemical nature and reactivity of the substance

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 :