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NeTakaya
3 years ago
11

A 2.00 L sample of gas at 35C is to be heated at constant pressure until it reaches a volume of 5.25 L. To what Kelvin temperat

ure must this sample be heated?
Chemistry
1 answer:
love history [14]3 years ago
7 0

Answer:

When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)

Explanation:

Step 1: Data given

The initial volume of the sample = 2.00 L

The initial temperature = 35 °C =  308 K

The increased volume = 5.25 L

Pressure = constant

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒ with V1 = the initial volume = 2.00 L

⇒ with T1 = the initial volume = 308 K

⇒ with V2 = the new volume = 5.25 L

⇒ with T2 = the new temperature

2.00 / 308 = 5.25 / T2

0.00649 = 5.25/T2

T2 = 5.25/ 0.00649

T2 = 808.9 K

When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)

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A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}

P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

#SPJ4

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