An ideal gas with 3.00 mol is initially in state 1 with pressure p1 = 20.0 atm and volume V1 = 1500 cm^3. First, it is taken to state 2 with pressure p2 = 1.50p1 and volume V2 = 2.00V1. Then it is taken to state 3 with pressure p3 = 2.00p1 and volume V3 = 0.500V1. What is the temperature of the gas in (a) state 1 and (b) state 2? (c) What is the net change in internal energy from state 1 to state 3?
Added by Ronnie Z.
Step 1
0 \, \text{atm}\) \(V_1 = 1500 \, \text{cm}^3\) \(n = 3.00 \, \text{mol}\) \(R = 0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K}\) Substitute the values into the formula: \(T_1 = \frac{P_1 \cdot V_1}{n \cdot R}\) Show more…
Show all steps
Your feedback will help us improve your experience
Prem Bijarniya and 74 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
An ideal gas initially at $300 \mathrm{~K}$ is compressed a constant pressure of $25 \mathrm{~N} / \mathrm{m}^{2}$ from a volume of $3.0 \mathrm{~m}^{3}$ to a volume of $1.8 \mathrm{~m}^{3}$. In the process, $75 \mathrm{~J}$ is lost by the gas as heat. What are (a) the change in internal energy of the gas and (b) the final temperature of the gas?
An ideal gas initially at $300 \mathrm{~K}$ is compressed at a constant pressure of $25 \mathrm{~N} / \mathrm{m}^{2}$ from a volume of $3.0 \mathrm{~m}^{3}$ to a volume of $1.8 \mathrm{~m}^{3}$. In the process, $75 \mathrm{~J}$ is lost by the gas as heat. What are (a) the change in internal energy of the gas and (b) the final temperature of the gas?
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD