What is the entropy change of the nitrogen if 250 mL of liquid nitrogen boils away and then warms to 20℃ at constant pressure? The density of liquid nitrogen is 810 kg/m³.

Question

Accepted Answer

Hello, fellow physicists today, we're to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A new diatomic gas called zoon is discovered zygon has a molar mass of 25.0 g per mole and a boiling point of °C, a 350 mL sample of liquid Zoon with a density of kg per meter cubed changes to gas at a constant pressure and reaches 45. °C. Z Sogo's latent heat of vaporization is 1.89 multiplied by 10 to the power of five joules per kilogram find the entropy change for Zoon. So that's our end goal is to find the entropy change of zoon. OK. So we're given some multiple choice answers. They're all in the same units of jewels per Kelvin. So let's read them off to see what our final answer might be. A is 1.09 multiplied by 10 to the power of three B is 2.51 multiplied by 10 to the power of three C is 3.27 multiplied by 10 to the power of four and D is 2.41 multiplied by 10 to the power of four. So first off, let us write down all of our known variables. So the volume of liquid for Zoon is leaders. But let's also write down what it is in meters cube. So when we convert it to meters cubed, it's multiplied by 10 to the power of six meters cubed. OK. And then the density of liquid Zoon, it's denoted as row is equal to 980 kilograms per meters cubed. And the boiling point of liquid Zoon, which is the initial temperature. In this case, it's gonna represent the initial temperature is negative because it's represented by the boiling point. So the initial temperature will be negative. So negative 171 °C. But let's convert that to Kelvin. So negative 171 plus equals 102 Kelvin. And then the final temperature of Zoon is 45 °C. Let's also convert that to Kelvin really fast. So 45 plus 273 is equal to Kelvin. And then the molecular mass of Zoon capital M subscript Z is 25 g per mole. And finally, the latent heat of vaporization is equal to 1.89 multiplied by 10 to the power of five Jews per kilogram. OK. So now let us note that initially the liquid zoon undergoes a phase transition to the gas state at a constant temperature which is an isothermal process. And then the temperature increases at a constant pressure which is an isobaric process. So both of these processes will increase entropy. The entropy change in an isothermal process is defined as delta S one is equal to Q divided by T I, the initial temperature where let's make a note here where Q equals the mass of Zoon multiplied by the latent heat of vaporization. So with that in mind, we can let's call it equation one, we can state that delta S one is equal to the mass of zoon multiplied by the latent heat of vaporization divided by the initial temperature. OK. So the entropy change at a constant pressure is defined as delta S two is equal to N multiplied by CP multiplied by Ln where Ellen is the natural log of TF the final temperature divided by the initial temperature. And let's call that equation two where N is the number of moles and CP is the specific heat capacity at a constant pressure. And let us also note that N the number of moles is equal to the mass divided by the molar mass which in this case, it's equal to MZ lowercase M subscript Z divided by capital M subscript Z. And let us also note that CP is equal to seven halves multiplied by capital R where capital R is the universal gas constant. So now we need to substitute the values for N and CP into equation two, let's call this equation three. So equations three states that delta S two is equal to MZ, lowercase MZ divided by capital MZ multiplied by seven halves multiplied by R multiplied by Ln of the final temperature divided by the initial temperature. So therefore the change in entropy of Zoon is delta SZ. So this is the change in entropy for Zoon is equal to delta S one plus delta S two. So now we need to substitute equations one and three into this equation. So delta SV is equal to MZ multiplied by LV divided by T I plus MZ divided by capital MZ multiplied by seven halves multiplied by R multiplied by Ln of TF the final temperature divided by the initial temperature. And let's call this equation four. OK. So now we need to find the mass of Zoon using the mass density formula. So the mass of Zoon is equal to the density multiplied by the volume which is equal to 980 kg per meter cubed multiplied by the volume which was 350 multiplied by 10 to the power of negative m cubed. The meters cubed canceled out, cancels out leaving us with 343 g when we multiply it in a calculator. So now we can solve for our final answer by plugging in all of our known variables into equation four. So delta SZ is equal to zero point kilograms, which if you're confused, we took 343 g and we use dimensional analysis on 1 kg. There is 1000 grams, the grams cancel out leaving us with kilograms which gives us 0.343 kg multiplied by 1.89 multiplied by 10 to the power of five Jews per kilogram, which is the latent heat of vaporization divided by the initial temperature which was 102 Kelvin. OK. Plus 343 grams divided by 25 g per mole multiplied by seven halves multiplied by and I'm running out of room. So I'm gonna bring it down below. So it's seven halves multiplied by the universal gas constant which is eight point jules per mole multiplied by Kelvin multiplied by Ln of final temperature which was Kelvin divided by 102 Kelvin the initial temperature. So when we take 0. kg multiplied by 1.89 multiplied by 10 to the power of five joules per kilogram. Divided by 102 plus g, divided by 25 g per mole multiplied by seven halves multiplied by the universal gas constant, multiplied by Ellen of the Ellen of the final temperature divided by the initial temperature. When we plug all of that into a calculator, we should get 1.09 multiplied by to the power of three jewels per Kelvin, which is our final answer. Ray we did it. So let's look at our multiple choice answers to see what the correct answer should be. It turns out that the correct answer is the letter A 1.09 multiplied by 10 to the power of three jewels per Kelvin. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

What is the entropy change of the nitrogen if 250 mL of liquid nitrogen boils away and then warms to 20℃ at constant pressure? The density of liquid nitrogen is 810 kg/m³.

Verified Solution

Key Concepts

Entropy

Phase Change

Specific Heat Capacity