An ideal-gas process is described by p=cV^ /2, where c is a constant. b. 0.033 mol of gas at an initial temperature of 150°C is compressed, using this process, from 300 cm^3 to 200 cm^3. How much work is done on the gas?

Video transcript

Hi, everyone in this particular practice problem, we're being asked to determine the work done by the gas. We will have the pressure of a gas sample following the relation of B equals to A V to the power of 0.5 where A is a constant. So if 0.12 moles of the gas is at a starting temperature of 20 degrees Celsius, expanding from 0.004 m cube to 0. m cube. Following the relation, we're being asked to calculate the work done by the guests. And the options given are a 463 Joles B, 100 and 37 Joles C negative 31.6 Jules and D negative 463 Joles. So we will assume that the gas is going to be an ideal gas. So let's just write that down as our assumption. And the work done on the gas is going to be given by the equation of W or work equals to negative integral of PDV. The P is the pressure and the V is the change in volume. So we are going to integrate this problem or this process from V zero to VF and the P or the pressure is going to follow the relation of P equals A V to the power of 0.5. So let's substitute that here. So P equals A V to the power of 0.5 multiplied by DV, just like. So, so we can actually integrate this or we are going to actually integrate this. But before that we want to remember or recall that the integration principle for X to the power of 0.5 multiplied by DX will be equal to one divided by 1. X to the power of 1.5 plus C. So in this case, let's use that integration principle in our integration here for the work done on the gas so that our work will then come out to be negative a multiplied by one divided by three divided by two or 1. multiplied by V to the power of three divided by two or 1.5. And we want to evaluate this from VF to or from V zero to VF and then simplifying this, we will get the word to then be equal to negative A multiplied by two divided by three V to the power of three divided by two or 1.5 evaluated from VF to V zero. And then we want to evaluate that so that our work will then come out to be two divided by three, multiplied by a multiplied by V zero to the power of three, divided by two minus VF to the power of three divided by two just like. So, so that will be the work done on the gas. So now we need to use the ideal gas law to solve for the constant A. And if you recall the ideal gas law will be P zero multiplied by P zero equals to nr multiplied by T zero. Solving for P zero, P zero will then come out to B through rearrangement and RT zero divided by P zero. So in this case, we are given the P to follow the equation of a multiplied by V to the power of 0.5. So let's equalize these two pressures where we'll have a multiplied by V zero to the power of 0.5 to be equals to N multiplied by R multiplied by T zero, divided by V zero. Rearranging uh Our equation that we obtain here for A, we can then obtain A to then be equal to N RT zero divided by V zero multiplied by V zero to the power of 0.5. So simplifying this A will equals to N RT zero divided by V zero to the power of three divided by two. We want to insert our A into the equation that we obtain for the work done so that the work done or W will then come out to be two divided by three. And let's substitute our A. So our A here equation will be substituted into our work done equation so that W will be equal to two divided by three multiplied by N RT zero, divided by 30 to the power of three divided by two multiplied by, in parentheses, P zero to the power of three divided by two minus VF to the power of three divided by two just like. So, so rearranging this and simplifying W will then come out to be two divided by three, multiplied by and RT zero multiplied by. We wanna um input the V zero to the power of 3.2 in the bottom here into our parentheses so that we will get one minus VF to the power of three divided by two divided by 30 to the power of three divided by two just like. So we can then actually substitute all of the information that we know given in the problem statement so that our W or where done will be equal to two divided by three multiplied by N which is 0.12 mole multiplied by R which is 8.31 Jules Burel Calvin multiplied by T zero which is going to be 293. Kelvin multiplied by in parentheses, one minus in parentheses. PF is 0.009 m cube divided by 0.004 m cube for P zero to the point to the power of three divided by two close parenthesis. And in calculating all of this, our W or the word done will then come out to a value of negative 463 Jews. So that will be the final answer to this particular practice problem with the work done by the gas as it is expanding will be equal to negative 460 s which will correspond to option D in our answer choices. So option D will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that'll be it for this video. And thank you so much for watching.