The cylinder in FIGURE CP18.73 has a moveable piston attached to a spring. The cylinder's cross-section area is 10 cm^2, it contains 0.0040 mol of gas, and the spring constant is 1500 N/m. At 20°C the spring is neither compressed nor stretched. How far is the spring compressed if the gas temperature is raised to 100°C?

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Accepted Answer

Hey, everyone. So this problem is dealing with the ideal gas law. Let's see what it's asking us. A cuboid shaped container with a movable piston contains 0.005 moles of gas at one atmosphere of pressure and 25 °C, the piston is attached to a spring with a spring constant of 2000 newtons per meter. If the piston is an equilibrium and the gas is heated to 75 °C, how much is the spring compressed? We are told to assume that the cross sectional area of the container is eight centimeters squared for multiple choice answers in units of centimeter are a 1.9 V, 0.9 C 0.5 or D 1.6. So when we're dealing with gasses under two different types of conditions, so one, we're at 25 °C and then we're at 75 °C. That's a good hint that we can use the combined gas law. So we can recall that that equation is P one multiplied by V one divided by T one is equal to P two multiplied by V two divided by T two. So when we look at this equation. Now, we don't know our initial volume, we don't know our final pressure and we don't know our final volume. We can also think about this spring force and cooks law. It gives us the equation F is equal to K that spring constant multiplied by delta X, that delta X is the change in spring length. And so we're asked for how much the spring is compressed. And so this term, this delta X is actually our target term for the problem. Now, we need to figure out how these two equations can be combined so that we're using the information from our gas to figure out that delta X. So we can recall that pressure by definition is the force over a given area. And so we can use that to eventually equate these two equations. So looking at our ideal gas law, which is, we can recall PV equals N RT, we do have everything for that first scenario, the one where we have the gas at 25 °C. So we can isolate the, on the left hand side of the equation. And that gives us N multiplied by R multiplied by T one all divided by P one. And so we are told our moles are 0.005 mo R is art ideal gas constant, which we can recall is 8.314 oops. And that's in units of Joel per Kelvin mole. And then our temperature 25 °C, we need to keep it in or put it into Kelvin. So that's 298 degrees Kelvin. And then all of that divided by our pressure, which was one atmosphere, which we can recall is 1.013 times 10 to the five pascals again, keeping everything in standard units. So we plug that into our calculators and we get a initial volume of 1.22 times 10 to the negative four cubic meters. When we think about what our second volume or V two will be, the spring will compress. And so our B two will be our initial volume plus that area multiplied by the change in spring length, right? How much that spring compresses. And so this is where we can see that delta X coming into the equation. Similarly, our P two, our pressure after the temperature is increased to 75 °C will be our initial pressure plus that force divided by area. After we have the, after that um temperature is increased, giving us the force from the spring. And so that will equal P one plus K multiplied by delta X divided by area. And so now we can combine these equations into our first combined gas, the law and we will get P one B one multiplied by T two, all divided by T one is equal to. So P two, which is P one plus K multiplied by delta X divided by a multiplied by V two, which is V one plus a multiplied by delta X. And so when we plug in these values, we will come up with, we can see it'll be a quadratic equation and then we can solve that. So P 11.013 times 10 to the fifth pascals multiplied by V one. We solve for 1.22 times 10 to the negative four cubic meters multiplied by T two. Again, keeping it in units of Kelvin. That's 348 degrees Kelvin. And then divided by T one is 298 degrees Kelvin. And that equals P one again 1.013 times 10 to the five pascals multiplied by or sorry plus plus our spring constant 2000 newtons per meter multiplied by delta X divided by our area that was given as eight centimeters squared. So that's 0.0008 m squared to keep it in standard units. And then V one again, 1.22 times 10 to the negative four cubic meters multiplied by area. I'm just gonna come down here 10008 multiplied by delta X. OK. When we plug all this in and expand the terms on the right hand side of the equation, we end up with 14.4. 3 Newton meters is equal to 12.36 Newton meters plus 386 multiplied by delta X plus 2000 multiplied by delta X squared. When we can recognize that that's a quadratic equation. We solve that for delta X and we get 0.005 m which is 0.5 centimeters. And so that's the final answer for this problem. And when we look at our multiple choice answers, we can see that aligns with answer choice C so that's all we have for this one, we'll see you in the next video.

Verified Solution

Key Concepts

Ideal Gas Law

Spring Force and Hooke's Law

Thermal Expansion and Gas Behavior