# First Law of Thermodynamics issue

• ghostops
In summary: So the equation becomes dU = nCp(Tf-Ti). Then you can solve for Tf using the values given and the ideal gas law.In summary, the conversation discusses an experiment involving 0.155 mol of air at a temperature of 780K and a pressure of 3.20x106 Pa in a cylinder of volume 310 cm3. 670 J of heat is transferred to the cylinder and the final temperatures of the air are calculated for two scenarios: when the volume is constant and when the pressure is constant. The equations used include dU= nCv(Tf-Ti) and dU = nCp(Tf-Ti), and the ideal gas law PV=nRT is used to solve for
ghostops

## Homework Statement

In an experiment to simulate conditions within an automobile engine, 0.155 mol of air at a temperature of 780K and a pressure of 3.20x106 Pa is contained in a cylinder of volume 310 cm3. Then 670 J of heat is transferred to the cylinder.

A) If the volume of the cylinder is constant while the heat is added, what is the final temperature of the air? Assume that the air is essentially nitrogen gas.

B) If instead the volume of the cylinder is allowed to increase while the pressure remains constant, find the final temperature of the air.

## Homework Equations

Const P W= pdV Q=nCpdT
Const V W = 0 Q = nCvdT
dU = nCdT
Cp = 5/2R
Cv = 3/2R

## The Attempt at a Solution

For part A where Volume is constant, it told me that 670 J of heat is added which would be the Q value for dU. so the overall internal energy is 670 J from that I used the equation dU= nCv(Tf-Ti) which with values came out too 670 = (.155)(3/2*8.314)(Tf-780) which equaled 1126 K. which ended up not being the right answer. I do not know what else to try.

For Part B, I know I need to figure out dU which is Q-W so for this equation it would come out to dU = 670 - pdV or dU = 670 - 3.20E6(Vf-3.1E-4m3). but don't really know where to go from here

• Nitrogen is a diatomic gas therefore Cv=2.5

For part B, find Vf as you have been finding and then use the ideal gas relation:

PV=nRT

the Diatomic gas worked, forgot about that tidbit.

but for the Part B I don't fully understand what you mean because I have 2 unknown variables V_f and T_f which is why I am a little lost.

Figured it out thanks!, took a good minute though

Can you please share the workings of the solution

For part b, you use the same equation and everything except that Cv is now Cp, which is just Cv + R.

## What is the First Law of Thermodynamics?

The First Law of Thermodynamics, also known as the Law of Conservation of Energy, states that energy cannot be created or destroyed, only transferred or converted from one form to another.

## How does the First Law of Thermodynamics relate to everyday life?

The First Law of Thermodynamics is applicable to everyday life in many ways. For example, it helps us understand how our bodies convert food into energy, how a battery stores and releases energy, and how a car's engine converts fuel into motion.

## What are some real-world examples of the First Law of Thermodynamics in action?

Some examples of the First Law of Thermodynamics in action include the process of photosynthesis, where plants convert sunlight into chemical energy, and the operation of a refrigerator, which transfers heat from inside the fridge to the outside environment.

## Why is the First Law of Thermodynamics important?

The First Law of Thermodynamics is important because it is a fundamental law of physics that helps us understand the behavior of energy in various systems. It also has practical applications in fields such as engineering, chemistry, and biology.

## How does the First Law of Thermodynamics relate to the Second Law of Thermodynamics?

The First Law of Thermodynamics and the Second Law of Thermodynamics are closely related. While the First Law states that energy is conserved, the Second Law states that in any energy transfer or conversion, some energy will always be lost as heat, making the total amount of usable energy decrease over time.

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