500 J of heat energy are transferred to a gas during a process in which the gas expands at constant pressure from 400 cm^3 to 800 cm^3 . The gas's thermal energy increases by 300 J during this process. What is the gas pressure?

Video transcript

Hey, everyone in this practice problem, we're being asked to calculate the pressure. We will have a gas enclosed in a piston container assembly losing 340 joles of heat as it is compressed from 450 centimeter cube to 200 centimeter cube at constant pressure. The internal energy of the gas increases by 220 jules. And we're being asked to calculate the pressure during the process. And the options given are a 2240 kg pascals B 480 kg pascals C 1000 kg pascals and D 480 kg pascals. So let's model the guest in the piston companion assembly as an ideal guess. So I'm just gonna write down ideal guess here. And we can then use the first law of thermodynamics where delta EPH will be equal to Q plus W. So first law of thermo where the delta eth or the change in internal energy will be equal to Q, which is the heat plus W which is the work done on the system of the gas. So in this practice problem, the gas will undergo isobaric compression at which it means that in the overall process, the gas will actually have a constant pressure. So isobaric compression where in the iso barry compression PF will be equals to B I will be equal to P. I'm just gonna represent it as P which is what we're interested to find. So therefore, the work done on the gas W can be obtained by negative P multiplied by delta V. So Delta Eth by substituting W here can be obtained by Q minus P multiplied by delta V. And we are given that Delta Eth will be equal to 220 Joes or the internal energy of the gas increases by 220 jos. So delta Eth equals to 220 Joles and Q will be equal to 340 Joles negative because it is going to be lost because the piston container assembly will lose 340 Joles of heat when it is compressed. And the delta V is obtained by VF minus V I which is going to be 200 centimeter cube minus 457 centimeter cube, which will come up to a value of negative centimeter cube or N si delta V will be equal to nega negative 250 times 10 to the power of negative 6 m cube. So we can then uh rearrange our Delta Eth equation that we have obtained here. So that B will then be equal to negative in parentheses. Delta eth minus Q close parenthesis divided by delta V. So now we can substitute all of our information to get the pressure where the pressure will be equal to negative in parenthesis. Delta Eth 220 S minus Q which is negative 340 Joles close parenthesis divided by delta V which is negative 250 times 10 to the power of negative 6 m Q. Calculating all of this, we will obtain the pressure B to B then equal to 2.24 times 10 to the power of six pascals or in kilo pascals. That will be 2240 kg pascals. So that will be the final answer to this particular practice problem where the pressure during the isobaric process is going to be kg pascals which will actually correspond to option A in our answer choices. So option A 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 will be it for this one. And thank you for watching.