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GAS LAWS

GAS LAWS. ASSUMPTIONS ABOUT GASES KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

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Page 1: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

GAS LAWS

Page 2: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

ASSUMPTIONS ABOUT GASES KINETIC THEORY OF GASES1. The volume of gas particles is so small

that they do not contribute to the volume in a container.

2. Particles of gas do not attract or repel each other

3. Particles of gas are in constant motion

Page 3: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

WHAT IS PRESSURE?

In your notes write down characteristics of what you think pressure is:

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BOOK DEFINITION ALL GASES EXERT PRESSURE Pressure: the force per unit area on a

surface

In other words, it’s how hard something is pushing against some sort of surface (for example a balloon or tire)

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UNITS OF PRESSUREAtmosphere (atm)Millimeters of mercury (mm Hg)Pascal (Pa)Conversions:

1atm = 760mm Hg = 101.3 kPa

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YOU HAVE TO BE ABLE TO CONVERT

Example: If you have 560 mm Hg, how many atmospheres is this? Use dimensional analysis Conversion factor 1 atm = 760 mm Hg

560mm Hg | 1 atm___ = 0.734atm | 760mm Hg

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TRY THESE1. If you start with 57.3 kPa, how

many atm is this?2. If you have 6.5 atm, how many

mm Hg is this?3. If you start with 435 mm Hg, how

many kPa is this?

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ANSWER

1. 0.566 atm2. 4.9 x 103 mm Hg3. 58.0 kPa

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DALTON’S LAW OF PARTIAL PRESSURES

Dalton’s Law of Partial Pressures: The total pressure of a system is equal to the sum of all the partial pressures of the gases in the system.

PTotal = P1 + P2 + P3 + . . .

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EXAMPLEYou have three gases in a container

(N2, H2, and CO2). If they have the following pressures that they exert:N2 = 360 mm HgH2 = 225 mm HgCO2 = 86 mm Hg

What is the total pressure in the container?

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ANSWER

PTotal = PN2 + PH2

+ PCO2

PTotal = 360mm Hg + 225mm Hg + 86mm Hg

= 671 mm Hg

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TRY THIS1. You have a container that has

the following gases (O2 + Cl2 + NO2). If the total pressure is 6.99atm and the PO2

is 2.25atm

and the PCl2 is 4.04atm, what is

the P NO2?

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ANSWERPT = PO2

+ PCl2 + PNO2

6.99atm = 2.25atm + 4.04atm + PNO2

PNO2 = 6.99atm – (2.25atm – 4.04atm)

PNO2 = 0.700atm

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BOYLE’S LAWTHOUGHT QUESTION:

If you squeeze a balloon (so that the volume is less), does the pressure inside the balloon increase or decrease?

Page 15: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

BOYLE’S LAWAs we all just saw, the pressure

increases.

If we were to increase the space of the balloon (without adding anymore air), would the pressure increase or decrease.

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BOYLE’S LAW The pressure would decrease

Therefore pressure and volume are inversely proportional

When pressure goes up, volume goes down and vice versa

The individual that noticed this relationship was Robert Boyle.

Pressure (P) = 1/Volume (V)

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THE RESULT Mathematical relationship of an

inverse proportion: PV = k (k is a constant that depends

on the environment . . . Temperature, number or

air particles, etc . . .) ASSUMPTIONS: Temperature

remains constant and the number of air molecules does not change

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BOYLE’S LAW If have the same conditions of

temperature and number of air molecules, then different pressures and volumes will still equal the same k:

P1V1 = k

P2V2 = k

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BOYLE’S LAWThe end result:

You can predict pressure or volume based on how you change the system:

P1V1 = P2V2

This equation is known as Boyle’s Law

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EXAMPLE If you start with a balloon that has

a volume of 3.5L and a pressure of 10 atm, what will be the new pressure if you squeeze the volume down to 1700mL?

Page 21: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

STEP 1Write down the information you

knowP1 = 10atm

V1 = 3.5L

P2 = ?

V2 = 1700mL

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STEP 2 Make sure that the units for volume and

pressure are the same. If not, then make them the same P1 = 10atm V1 = 3.5L P2 = ? V2 = 1700mL = 1.7L

L and mL are not the same, so we convert mL to L to make them the same unit

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STEP 3Put the values in the formula

P1V1 = P2V2

(10atm)(3.5L) = P2 (1.7L)

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STEP 4Solve the equation so that the units

cancel out(10atm)(3.5L) = P2 (1.7L)

P2 = (10atm)(3.5L)

1.7LP2 = 21 atm

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TRY THESE1. If you have a container that has a

volume of 25.0L and a pressure of 760mmHg, what is the new volume if you increase the pressure to 2.40atm?

2. If you start with a volume of 250mL and a pressure of 67.3kPa and change the volume to 1.00L, what is the new pressure?

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SOLUTION

1. 10.4L2. 16.8 kPa

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CHARLE’S LAW Thought question:

What happens to air as it is heated? Think of a hot air balloon

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CHARLE’S LAWAnswer: As you heat air, or

any gas, it causes the volume to increase

In addition, if you cool air, the volume of the gas will decrease

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CHARLE’S LAW Therefore, temperature and volume are

directly proportional As temperature increases, volume

increases As temperature decreases, volume

decreases The individual who noticed this

relationship was Jacques Charles Volume (V) = Temperature (T)

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CHARLE’S LAWMathematical relationship of a

direct relationship:V/T = k (k is the same constant as

before)ASSUMPTIONS: That the system

has a constant pressure and the same number of gas particles.

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CHARLE’S LAW If have the same conditions of

pressure and number of air molecules, then different temperatures and volumes will still equal the same k:

V1/T1 = kV2/T2 = k

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CHARLE’S LAW The end result:

You can predict temperature or volume based on how you change the system:

V1/T1 = V2/T2

This equation is known as Charle’s Law

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REVIEW OF KELVIN TEMPERATURE

Before we go any further, we need to review a special unit of temperature: Kelvin (K)

A temperature scale based on absolute 0 (the coldest possible temperature)

The Kelvin temperature converts to Celsius: K = °C + 273

Page 34: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

KELVIN TEMPERATURE

In any of the gas laws, you MUST first convert all temperatures into K before you can use the formulas

Page 35: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

EXAMPLE If you have a container that has a

temperature of 25 °C and a volume of 3.4L, what would be the new temperature if you increased the volume to 5.0L?

Page 36: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

STEP 1Write down your information

T1 = 25°C

V1 = 3.4L

T2 = ?

V2 = 5.0L

Page 37: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

STEP 2 Make sure that the units for volume are

the same and temperature is in kelvins (K). If not, then convert. T1 = 25°C + 273 = 298K V1 = 3.4L T2 = ? V2 = 5.0L

Converted Celsius to kelvin and the volume has the same unit already.

Page 38: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

STEP 3 Put the values in the formula

V1/T1 = V2/T2

(3.5L)/(298K) = (5.0L)/T2

NOTE: This is where you cross multiply

Page 39: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

STEP 4 Solve the equation so that the units

cancel out

(3.5L)/(298K) = (5.0L)/T2

T2 = (5.0L)(298K)/(3.5L)

T2 = 426K

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TRY THESE1. A container has a temperature of 30°C

and a volume of 335mL. If you heat the container to 50 °C, what is the new volume?

2. A balloon has a volume of 25.0L and a temperature of 0 °C, what is the new volume if you decrease the temperature to -25 °C?

Page 41: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

SOLUTION

1. 357 mL2. 22.7L

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TRY THISA box has a temperature of -25°C and the dimensions of 25cm x 10cm x 0.5cm. If you heat the container to 280 K, what is the new volume?

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SOLUTIONFind the following:

V1= 25cm x 10cm x 0.5cm = 125mLT1= -25°C + 273 = 248KV2= ?T2= 280K

V2 = 141mL

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COMBINED GAS LAWThere is a way of looking at all three conditions of:

TemperatureVolumePressure

at the same time

Page 45: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

COMBINED GAS LAWLet’s look at Boyle’s Law

PV = kLet’s also look at Charles’ Law

V/T = kTHOUGHT QUESTION:

How do you think you could combine P, V, and T to equal k in one equation?

Page 46: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

COMBINED GAS LAWNotice:

P and V are multiplied together (PV) V is divided by T (V/T)

If you put them together: PV/T = k

ASSUMPTIONS: You have the same number of air molecules.

Page 47: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

COMBINED GAS LAW If have the same number of air

molecules, then different temperatures, volumes and pressures will still equal the same k:

P1V1/T1 = k

P2V2/T2 = k

Page 48: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

COMBINED GAS LAW The end result:

You can predict temperature, volume or pressure based on how you change the system:

P1V1/T1 = P2V2/T2

This equation is called the combined gas law

Temperature (T) must always be in Kelvin (K)

Page 49: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

EXAMPLEA container has a starting

pressure of 2.50atm, a starting temperature of 33°C and a starting volume of 1250 mL. If you change the temperature to 10 °C and the volume to 1.0L, what is the new pressure?

Page 50: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

STEP 1Write down your information

P1 = 2.50atmV1 = 1250mLT1 = 33 °CP2 = ?V2 = 1.0LT2 = 10 °C

Page 51: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

STEP 2 Make sure that the units for volume and

pressure are the same and temperature is in kelvins (K). If not, then convert. P1 = 2.50atm V1 = 1250mL = 1.250L T1 = 33 °C + 273 = 306K P2 = ? V2 = 1.0L T2 = 10 °C + 273 = 283K

Page 52: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

STEP 3 Put the values in the formula

P1V1/T1 = P2V2/T2

(2.50atm)(1.250L)/306K = P2(1.0L)/283K

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STEP 4 Solve the equation so that the units cancel

out (2.50atm)(1.250L)/303K = P2(1.0L)/283K

P2 = (2.50atm)(1.250L)(283K)/(303K)(1.0L)

P2 = 2.9atm

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TRY THISA container has a volume of

2500mL, a pressure of 700 mmHg and a temperature of 15 °C. If you change the pressure to 2.5atm and you change the temperature to 300K, what is the new volume?

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SOLUTION

9.6 x 102 mL

or 0.96L

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IDEAL GAS LAW How does an ideal gas behave?1. The volume of gas particles is so small

that they do not contribute to the volume in a container.

2. Particles of gas do not attract or repel each other

3. Particles of gas are in constant motion

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IDEAL GAS LAWWhen we make these

assumptions, we can create a universal IDEAL GAS LAW

PV = nRT

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VARIABLES P = pressure (must be in the units of atm) V = volume (must be in the units of L) n = moles T = temperature (must be in the units of

K) R = gas constant

R = 0.0821 L*atm/mol*K only works with ideal gases

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EXAMPLEA gas is enclosed in a box that

is 5.4L in volume. If you have 22.4 moles at a temperature of 22°C, what is the pressure exerted on the box?

Page 60: GAS LAWS. ASSUMPTIONS ABOUT GASES  KINETIC THEORY OF GASES 1. The volume of gas particles is so small that they do not contribute to the volume in a

STEP 1Write down your information

P = ?V = 5.4Ln = 22.4 molesR = 0.0821 L*atm/mol*KT = 22°C

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STEP 2 Make sure that each measurement has

the correct units. P(atm), V(L), n(moles) and T (K). P = ? V = 5.4L n = 22.4 moles R = 0.0821 L*atm/mol*K T = 22°C = 295K

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STEP 3 Put the values in the formula

PV = nRT

(P)(5.4L) = (22.4moles)(0.0821 L*atm/mol*K)(295K)

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STEP 4 Solve the equation so that the units

cancel out

• (P)(5.4L) = (22.4moles)(0.0821 L*atm/mol*K)(295K)

• (P)(5.4L) = 542 L*atm 5.4L 5.4L• P = 1.0 x 102 or 100 atm

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TRY THISA container has a volume of

2500mL, a pressure of 3.5 atm and a temperature of 25 °C. If this is an ideal gas, how many moles do you have?

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SOLUTION

0.36 moles

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MORE ON ASSUMPTIONSFor each of the gas laws to work

we have to follow certain assumptions . . . just like the Ideal Gas Law

With these assumptions we can convert the ideal gas law into the other gas laws

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ASSUMPTIONS Boyle’s Law

Charles’ Law

Combined Gas Law

Temperature remains constant and the number of moles does not change

That the system has a constant pressure and the number of moles does not change.

You have the same number moles.

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CONVERTING THE IDEAL GAS LAW

We start with the ideal gas law PV = nRT

We look at the assumptions of each law to see what must remain constant If the condition remains constant, we

can remove it from the ideal gas law We set the equation equal to the gas

constant (R) to get the other gas laws EXAMPLES ON THE BOARD

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STOICHIOMETRY REVISITED

With gases, we sometimes work at a state called STANDARD TEMPERATURE AND PRESSURE (STP).

The values of STP are:Pressure = 1 atmTemperature = 273K (0°C)

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VOLUME If you have one mole of a gas at

STP, what is the volume?

TRY IT OUT

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VOLUME AT STPGiven:

Pressure = 1 atmVolume = ?n = 1 moleR = 0.0821 L*atm/mol*K T = 273 K

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ANSWER

1 mole of gas = 22.4L at STP

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NOW WE CAN USE THIS IN DIMENSIONAL

STOICHIOMETRYTry the following:

You begin with 9.66x1033 molecules of nitrogen gas. What is the volume at STP?

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SOLUTION Just like you did previously, you convert

molecules to moles Since you know 1 mole = 22.4L at STP,

that is the second step

9.66x1033 mq | 1 mole | 22.4L | 6.02x1023 mq | 1 mole

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ANSWER

3.85x103.85x1011 11 LLNotice: It doesn’t matter

what gas you have, they are all 22.4 L/mole

Therefore, you don’t need the density to solve for the volume of a gas

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TRY THISYou have 336mL of O2 gas at

STP. How many oxygen molecules do you have?

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ANSWER

9.03X1021 molecules of O2

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STOICHIOMETRYSince we do not need density to solve for volume, we can also solve normal stoichiometry problems with gasesIF THEY ARE AT STP

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TRY THIS Translate and balance:

Hydrogen gas is combined with nitrogen gas to form ammonia (NH3) gas.

If you start with 25g of nitrogen gas, what volume of ammonia gas do you create?

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3H2 + N2 2NH3

25g N2 | 1 mole N2 | 2 moles NH3 | 22.4 L NH3__

| 28 g N2 | 1 mole N2 | 1 mole NH3

= 40L

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WHAT DO YOU USE?A sample of nitrogen gas has

a mass of 55.4g. If the gas is at STP, what is the volume that the gas occupies?

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MORE CHALLENGINGA sample has a volume of

250mL at a temperature of 25°C. If this sample is composed of 0.50g of H2, what is the pressure of the sample?

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WHAT DO YOU USE?

A container has a volume of 254mL and a temperature of 25°C. If you increase the temperature to 373K, what is the new volume?

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WHAT DO YOU USE?A system has a volume of

550mL and a pressure of 225kPa. If you have 32.5g of chlorine gas, what is the temperature?

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WHAT DO YOU USE?A sample of gas has a pressure of

1250 mmHg in a volume of 3.55L. If you change the pressure to 295kPa, what is the new volume?

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WHAT DO YOU USE?A container contains a gas that

has a pressure of 2.95atm, a volume of 455mL and a temperature of 15°C. If you change the pressure to 700mm Hg and the temperature to 35°C, what will be the new volume?