Sign up for our free STEM online summer camps starting June 1st!View Summer Courses

### Limestone stalactites and stalagmites are formed …

03:01
Manhattan College

Need more help? Fill out this quick form to get professional live tutoring.

Get live tutoring
Problem 102

A sample of gas is contained in a cylinder-and-piston arrangement. It undergoes the change in state shown in the drawing. (a) Assume first that the cylinder and piston are perfect thermal insulators that do not allow heat to be transferred. What is the value of $q$ for the state change? What is the sign of $w$ for the state change? What can be said about $\Delta E$ for the state change? (b) Now assume that the cylinder and piston are made up of a thermal conductor such as a metal. During the state change, the cylinder gets warmer to the touch. What is the sign of $q$ for the state change in this case? Describe the difference in the state of the system at the end of the process in the two cases. What can you say about the relative values of $\Delta E ?$

(a)$q=0, w$ is positive, and $\Delta E$ increases.

## Discussion

You must be signed in to discuss.

## Video Transcript

Okay, guys. Running problem number one dring to in chapter five of chemistry of the Central Science. So a sample of gas it contained in a cylinder and piston arrangement it undergoes the change in st shown in in the drawing, you can see the centre textbook assume first of the cylinder piston are perfect normal insulators that you allow heat to be transferred. What is the value of Q for? To stay changed. So cue for apart A. That's going to be zero If the thermal insulator and know he's transferred, then you're qu zero. What is the sign of the the state of it? Because I am w for the state change w is equal to I gotta pee Delta V Since Delta V is he is less than zero because you're going from high volume too low volume you're you're seeing compression vision w is going to be greater than zero. So W is going to be positive talking about lto even the reaction afraid to stay change. So Delta is equal to que plus w But since Q zero Delta is going to be positive because it is going to equal work, No, okay, for part B, they say. Now seen this cylinder piston are made up of a thermal conductors such as metal. There, any state change to sell that gets warm to touch. Well, it's a sign of Cuban this reaction from this for this state. So if Dean warmer to the touch queues as that means he's being released, so Q is going to be less than zero. So he was going to be negative of now. Describe different state of the system with the end of processed through cases. What do you say about the relative value of of Delta E. So in Delta, that's easy for this case is going to be equal Tio que plus w Now the work you do in both cases, that's going to be exactly the same and how, because you're compressing it. The volume is still volume and change volume. So constant now. Oh, because Q is negative here. The is going to your when you add it to your do you work here you're the value that you would get from. That sum is going to be lower than the value ofwork itself, because if you add a positive plus a negative number, you get answers going to be less than you're positive. So that means that less label this Delta E two and this one Delta one that means Delta E two is going to be Delta. One is going to be greater than Delta E to. It's going to be more positive. We can't tell for sure how much more positive it is, but we but we can know that it's it has a larger value.