Chemistry College

A sample of gas at a constant
volume initially has a
temperature of 315.0 K with a
pressure of 2.50 atm. The
pressure changes to 3.61 atm.
Calculate the final temperature.

Answers

Answer 1

The final temperature of the gas, when the pressure changes from 2.50 atm to 3.61 atm at constant volume, is approximately 454.9 K.

What is the final temperature of the gas?

Gay-Lussac's law states that the pressure exerted by a given quantity of gas varies directly with the absolute temperature of the gas.

It is expressed as;

P₁/T₁ = P₂/T₂

Given that:

Initial pressure P₁ = 2.50 atmInitial temperature T₁ = 315.0 KFinal pressure P₂ = 3.61 atmInitial temperature T₂ = ?

Plug the given values into the equarion above.

P₁/T₁ = P₂/T₂

P₁T₂ = P₂T₁

[tex]T_2 = \frac{P_2T_1}{P_1}\\ \\T_2 = \frac{3.61 \ *\ 315.0}{2.50} \\\\T_2 = 454.9K[/tex]

Therefore, the final temperature is 454.9 K.

Learn more about Gay-Lussac's law here: brainly.com/question/1358307

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Related Questions

Calculate the change in heat when 16.00 g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 25.50 °C.

Answers

The change in heat when 16.00 g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 25.50°C is -31.25 kJ (exothermic process).

To calculate the change in heat when 16.00 g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 25.50 °C, we need to consider two separate processes and add their heat changes together:

The heat change during the condensation of steam to liquid water:

The heat change during this process can be calculated using the heat of vaporization of water, which is 40.7 kJ/mol at 100.0°C.

First, we need to determine the number of moles of water in 16.00 g:

moles of water = mass of water / molar mass of water

moles of water = 16.00 g / 18.015 g/mol

moles of water = 0.8886 mol

The heat change during the condensation of steam to liquid water can be calculated as follows:

q1 = moles of water x heat of vaporization

q1 = 0.8886 mol x 40.7 kJ/mol

q1 = 36.21 kJ

The heat change during the cooling of liquid water from 100.0°C to 25.50°C:

The heat change during this process can be calculated using the specific heat capacity of water, which is 4.184 J/g°C.

The temperature change during this process is:

ΔT = final temperature - initial temperature

ΔT = 25.50°C - 100.0°C

ΔT = -74.50°C

The heat change during the cooling of liquid water can be calculated as follows:

q2 = mass of water x specific heat capacity x ΔT

q2 = 16.00 g x 4.184 J/g°C x (-74.50°C)

q2 = -4,958 J

Therefore, the total heat change for the two processes is:

ΔH = q1 + q2

ΔH = 36.21 kJ + (-4,958 J)

ΔH = 36.21 kJ - 4.958 kJ

ΔH = 31.25 kJ

Therefore, the change in heat when 16.00 g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 25.50°C is -31.25 kJ (exothermic process).

Learn more about water vapor at

brainly.com/question/14529481

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The heat of vaporization for ethanol is 0.826 kJ/g
. Calculate the heat energy in joules required to boil 75.25 g
of ethanol.

Answers

Answer:

87469.73J

Explanation:

72.25g/0.826kJ/g=87.4697337kJ

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