in this video we're going to focus on chemistry problems related to internal energy heat and work so let's start with this one calculate the change in the internal energy of a system if 300 joules of heat energy is absorbed by the system and if 400 joules of work is done on the system now i don't know if you saw a previous video that i created on the first law of thermodynamics but if you haven't in chemistry delta u the change in internal energy is equal to q plus w in physics it's q minus w now q is positive whenever heat is absorbed by the system so that's during an endothermic process q is negative whenever heat is released by the system so that's during an exothermic process w is positive whenever work is done on a system so the internal energy of the system will go up and w is negative whenever work is done by the system so this would decrease the internal energy of the system so those are a few things to keep in mind throughout this video now let's get back to this problem what is the value of q and what is the value of w now notice that the system absorbs 300 joules of heat energy because it absorbs energy q is positive now notice that work is done on a system when work is done on a system w is positive so work is going to be positive 400 joules now using that equation delta u is q plus w so we have 300 joules of heat energy absorbed by the system plus 400 joules of work is done on a system so both of these events work to increase the internal energy of the system so the change in the internal energy is 700 so if you want to draw a picture let's say inside the box represents the system and everything outside of that is the surroundings so the system absorbs 300 joules of energy so the system gains 300 which means the surroundings loses 300 joules so this process is endothermic for the system but exothermic for the surroundings now work is done on a system 400 joules of work is done on the system so the system gains 400 joules of energy but the surroundings loses 400 joules of energy through work so we could say work is done on a system but work is done by the surroundings let's try this one the system releases 700 joules of heat energy and 300 joules of work is done by the system calculate the change in the internal energy of the system so let's start with a picture first so this is going to be the system and outside of that we have the surroundings so the system releases 700 joules of heat energy so 700 joules of heat energy transfers out of the system so this is going to be negative 700 for the system because it lost that energy but the surroundings gain 700 joules of heat energy now 300 joules of work is done by the system if the system is doing work it's expending energy to do that so the system is going to lose another 300 joules of energy but the surroundings will gain that 300 joules of energy so the surroundings gain a total of 1 000 joules but the system loses a total of a thousand joules and so we have the first law of thermodynamics energy is neither created or destroyed is simply transferred from one place to another so delta u i forgot the u which is q plus w it's going to be negative 700 plus negative 300 so the change in the internal energy of the system is negative one thousand so if the system loses a thousand joules of energy the surroundings gain a thousand joules of energy but this is the answer to the problem number three what is the change in the internal energy of the system if the surroundings gain 250 joules of heat energy and if 470 joules of work is performed by this romans so we need to visualize the transfer of energy so if the surroundings gain 250 joules of energy is that heat energy flowing into the system or into the surroundings that energy is flowing into the surroundings if it flows if the surroundings gain that energy so therefore we could say q with respect to the system is negative 250 joules because heat energy is coming out of the system going into the surroundings now 470 joules of work is performed by the surroundings whenever something performs work it loses energy to do so for example if you perform the work required to lift weights you have to burn energy to do it so if the system is doing work the system is losing energy so energy is transferring i mean if the surroundings is doing work the surroundings loses energy which means energy is transferred from the surroundings to the system so we got 470 joules of work leaving the surroundings going to the system so w is positive 470. so delta u which is q plus w it's negative 250 plus 470 so that's 220 so i'm going to clarify if work is performed by the surroundings the surroundings is losing energy due to work and that energy is going into the system which means work is being done on a system if it's done by the surroundings and anytime work is done on a system w is positive number four what is the change in the internal energy of the system if the surroundings releases 300 joules of heat energy and if the system does 550 joules of work on the surroundings so go ahead and try this problem now if the surroundings releases 300 joules of heat energy what does that mean well that means the surroundings is losing energy so the system is gaining that energy so if the system absorbs 300 joules of heat energy q is positive 300. now the system does 550 joules of work on the surroundings so work is being done by the system but on a surroundings anytime work is done by the system work is negative energy is flowing out of the system to the surroundings so whenever work is done by something energy is being consumed so if work is done by the system the system loses energy as it's doing work it's expended energy and if work is done on its surroundings the surroundings is gaining energy so make sure you understand what these expressions mean so now let's calculate the change in the internal energy of the system so it's q plus w so the system gains 300 joules as you can see that's flown into the system but it's losing 550 joules you can see that's flowing out of the system so then that result is that the system is losing 250 joules of energy so delta u is negative we got a net energy flow out of the system number five how much work is performed by a gas as it expands from 25 liters to 40 liters against a constant external pressure of 2.5 atm so what equation should we use for a problem like this well let's draw a picture so this time the system is a gas so let's say this cylinder is filled with gas particles so that's the system and outside we have the surroundings now the surroundings exerts a force on this gas anytime you have a pressure there's a force pressure is force divided by area so there's a constant external pressure of 2.5 atm and that pressure exerts a force in the gas as a result the gas is going to compress and this problem is going to expand but we'll talk about that later right now i'm just deriving the formula so as we apply a force on a gas the gas will compress so the volume is being reduced and notice the change in the height of the cylinder that's delta h now to calculate the work done by a gas or on a gas in this case it's on a gas it's force times the displacement which in this case the displacement in the y direction is the change in height and based on this equation if you rearrange it if you multiply both sides by a force is pressure times area now the volume of a cylinder is the area times the change or times the height so area times height is the volume of the cylinder the area is the area of this circle which is pi r squared so this is the volume of the cylinders pi r squared times height so if the volume is area times height then area times the change in height must be the change in volume now we need to add a negative sign to make this work due to the sign conventions now during compression as was the case relating to the picture that we have the change in volume is negative the volume is decreasing and a force was being applied on the gas in order to compress it so the surroundings was doing work on the gas and whenever work is done on the gas or on the system w is positive during expansion whenever a gas expands the volume increases and the gas is doing work on the surroundings as it expands and so work is going to be negative so during compression let me see if i could draw this picture here you're applying a force to compress a gas so you're doing work on the gas which means you're increasing the internal energy of the gas whenever the volume decreases the pressure increases based on boyle's law so as you apply a force to compress a gas what you're really doing is you're expanding energy and that energy that you're expending it's being stored in the form of pressure so whenever you compress a gas you're transferring energy to that gas you're increasing its pressure and whenever that gas decides to expand it's going to apply an upward force and as it applies in upward force it can do work on the surroundings so as the volume expands the pressure reduces so in order to store energy in a gas you need to compress the gas and to release that energy the gas has to expand and keep in mind the energy is stored in the form of pressure which is a type of potential energy so gases that have a high pressure has a lot of stored energy gas is at low pressure doesn't have much stored energy so keep that in mind so if you're doing work on the gas you got to apply a force so energy is being transferred from you to the gas now as the gas expands against the surroundings energy is flowing from the gas to the surroundings so during compression w is positive work is done on the gas during expansion w is negative work is done by the gas and so the internal energy decreases during expansion but it increases during express compression now notice that delta v and w always have opposite signs so based on that w is negative p delta v so when delta v is negative you're going to have two negative signs which means w has to be positive as in the case of this problem or during expansion when delta v is positive w has to be negative because a negative times a positive number equals a negative number and so because these two signs are opposite that's why we have the negative sign in front of the equation so just make sure you understand that now let's focus on a problem at hand how much work is performed by a gas as it expands from 25 liters to 40 liters against a constant external pressure of 2.5 atm so we said the equation is negative p delta v the change in volume is the final volume minus the initial volume now this pressure is not the internal pressure of the gas because that gradually changes from 4 atm to 2.5 atm p represents the constant external pressure that the gas has to work against which was the 2.5 atm the final volume is 40 the initial volume is 25 so during expansion delta v is positive and we said w has to be negative whenever a gas expands so 40 minus 15 i mean 40 minus 25 is 15 and 2.5 times 15 that's 37.5 so the work is negative 37.5 liters times atm because the volume was in liters and the pressure was an atm so anytime a gas expands the work done by the gas is negative now you need to be able to convert this answer to joules and here's the conversion that you need one liter times one atm is equal to 101.3 joules so w is three thousand seven hundred and ninety nine joules if you round it to a nearest whole number and don't forget this is negative number six how much work is required to compress a gas from 50 liters to 35 liters at a constant pressure of 8 atm so during compression is the work going to be positive or is it going to be negative to compress a gas the work done on a gas is positive the internal energy of the gas will increase so let's go ahead and use this formula it's negative p delta v so the pressure is constant it's 8 atm and delta v is the final volume which is 35 liters minus the initial volume of 50 liters so what we have is negative 8 atm multiplied by a change in volume of negative 15 liters so therefore the work required is positive 120 liters times atm so now let's convert this to joules so recall that one liter times one atm is equal to 101.3 joules so these units cancel so it's 120 times 101.3 and you should get 12 156 joules so that's the work required to compress the gas from 50 liters to 35 liters at a constant pressure of 8 atm number seven 500 joules of heat energy was absorbed from the surroundings and the gas expanded from 30 liters to 70 liters against the constant pressure of 2.8 atm calculate the internal energy change in joules so let's start with this equation w is negative p delta v the change in volume is equal to the final volume minus the initial volume now the pressure is 2.8 atm and the volume the final volume is 70 liters minus the initial volume of 30 liters so 70 minus 30 is 40 so we have negative 2.8 atm multiplied by positive 40 liters so the change in volume is positive due to the expansion of the gas which means work has to be negative so negative 2.8 times 40 is negative 112 with the units being liters times atm in the last problem w was 120 and that was liters times atm i didn't convert it to joules but you know how to convert it to joules in this problem we need to because we want the final answer to be in joules so i'm going to multiply this answer by 101.3 joules per liter per atm so that these units will cancel so it's negative 112 times 101.3 so the work due to the expansion of this gas is negative 11 345.6 joules so now we need to calculate delta u what is q is q positive 500 or negative 500. so notice that heat energy was absorbed from the surroundings if it's absorbed from the surroundings that means heat flows from the surroundings to the system so heat energy is absorbed by the system so q is positive 500 the system gains 500 joules from the surroundings the system being the gas by the way so now delta u is q plus w so that's 500 plus negative 11 345.6 so the change in the internal energy of the system is negative 10 845.6 joules so this is the answer you