A helium balloon is filled to a volume of 27.7 L at 300 K.?
PV = n * R * T P[1] * V[1] = n * R * T[1] P[2] * V[2] = n * R * T[2] (P[2] * V[2]) / (P[1] * V[1]) = n * R * T[2] / (n * R * T[1]) (P[2] * V[2]) / (P[1] * V[1]) = T[2] / T[1] P[2] * V[2] * T[1] = P[1] * V[1] * T[2] V[1] = 27.7 L V[2] = ? T[1] = 300 K T[2] = 392 K Presuming that the pressure remains constant, P[1] = P[2], so the formula simplifies P[2] = P[1] = P P * V[2] * T[1] = P * V[1] * T[2] V[2] * T[1] = V[1] * T[2] V[2] = V[1] * T[2] / T[1] V = 27.7 * 392 / 300 V = 36.1946666667 V = 36.2 Liters
A gas-filled weather balloon with a volume of 58.4 L is released at sea-level conditions of 755 torr and 23°C.
okay this is very simple. for this question you are solving for V2 = volume 2 or final volume. 835L is not the final volume you will be solving for, thats the maximum volume that the balloon can hold okay. initial condtions: Volume = 55.0 L Pressure = 755 torr or 0.9934 atm temperature = 23 degrees celcius or 296.15 kelvin final conditions: Volume = ? pressure = 0.066 atm temperature = -5 degrees celcius or 268.15 kelvin remember the question is asking you if at the final conditions is the volume of the balloon at the max or not. okay and you will simply just use the general gas equation (P1 * V1) / t1 = (P2 * V2) / t2 >>> where 1 refers to the initial conditions and 2 refers to the final conditions and P = pressure, V = volume and t = temperature. okay so now all you have to do is isolate for V2 and that would be... V2 = (P1 * V1 * t2) / (t1 * P2) = (0.9934atm)(55.0L)(268.15k) / (296.15k)(0.066atm) = ANSWER!! sorry i dont have a calculator on me, but this is correct because if you do the unit analysis ( see which units cancel out) you are left with L which is the unit for volume!. so anyways if the volume is 835L then yes it will reach its maximum volume at the final conditions and if its not 835L then no.
A rubber balloon is filled with 1 L of air at 1 atm and 300 K and is then cooled to 100 K?
Use the combined gas law equation relating temperature, pressure, and volume together. It's P1V1/T1 = P2V2/T2 Where p is pressure, v is volume, and T is temperature. I'm going to assume the final pressure is going to be 1 atm because you haven't specified otherwise. So (1 atm)(1 L)/300K = 0.0033 That means whatever your final crap is, it has to equal 0.0033 or close enough to be acceptable. V2 x 1atm/100 K = 0.0033 Rearrange that 0.0033 x 100/1 atm = V2 0.33 = V2 Plug it back in and you get 0.0033 Check your answer 0.0033 vs. 0.0033 Ta da! You know the math is right. So the final volume actually decreased by 0.67 liters. That means that as the temperature decreases, the volume must decrease as well. I have proved the combined gas law does indeed work. ;)
Assuming you want to only increase the temperature of the 'air' in the room (and not the walls and other things), the calculation will use the specific heat capacity of air, the mass of air and the temperature difference to arrive at the amount of energy needed.Assuming the height of the room is about 10 ft tall, the total volume of the room is 10 x 300 = 3000 cu.ft.Amount of energy needed to raise temperature is given by the formula: specific heat capacity of substance x mass of substance x temperature difference1. Specific heat capacity of air is 1.006 kJ/kgC2. mass of air: density of air is 0.036 kg/cu.ft. mass = density x volume = 0.037 x 3000 = 111KG3. Temperature differential = 70C - 60C = 10Cenergy needed = 1.006 x 111 x 10 = 1116 KJPower is energy transferred / second. That means, your 1000W space heater can transfer 1000J of energy / second. So time taken to heat up this space will be:1116000 / 1000 = 1116 seconds. In other words: 18.6 minutes.This of course assumes that the output power of your heater is 1000W. If the 'input' is 1000W, you may want to find out the efficiency of your heater and then arrive at the actual output power of the heater by the formula output power = input power x efficiency.
A balloon is filed with helium gas at 20 degree celsius and 1.0 atm pressure?
First law of thermodynamics states ΔU = Q – W, where ΔU is the change in internal energy, Q the heat gained (or lost) by the helium gas and W the work done by the gas. From thermodynamics of gasses follows ΔU = nC(v) x ΔT, where C(v) = 3/2 R = molar specific heat for helium (monoatomic gas), so ΔU = 3/2 nR x ΔT. We need the number of moles n for helium, which we find from ideal gas law: p1 x V1 = n R T1 or n = p1 x V1 / R x T1 and ΔU = 3/2 x p1 x V1 x ΔT / T1 = 3/2 x 1.01 x 10^5 N/m² x 8.50 m³ x 35 K / 293 K = 153,827 J. W = p1 x ΔV (the gas is expanding, so it does work by pushing the surrounding air). We need the change in the volume of the balloon, which we can obtain from Charles law (isobaric process, p = const.): V1/T1 = V2/T2 or V2 = V1 x T2 / T1 = 8.50 m³ x 328 K / 293 K = 9.52 m³ and ΔV = V2 – V1 = 1.02 m³, so the work done is W = 1.01 x 10^5 N/m² x 1.02 m³ = 103,020 J. So Q = ΔU + W = 153,827 J + 102,020 J = 154 kJ.
Recall the Combined Gas Law:P1V1/T1 = P2V2/T2In this case, at STP, P1 = 1atm, V1 = 2x22.4L, T1 = 273K (0C)P2 = 0.98atm, V2 is unknown, T2 = 310K (37C)So, 1x44.8/273 = 0.98xV2/310Solving for V2, we find V2 = 51.91LSo the volume of the balloon is 51.91L.
According to Wikipedia: hydrogen has a density of 0.08988 g/L, oxygen has a density of 1.429 g/L, and air has a density of 1.2922 g/L at 0deg C and 101.325 KPa.Let’s say both balloons are the same size, one liter. Together, they will weigh 1.51888 g, and will displace two liters of air, which weighs 1.5844 g. This means the balloons together will have a net positive buoyancy.But if the oxygen balloon is larger than the hydrogen balloon, this won’t be the case. Let’s say that both balloons together contain one liter, so they will displace one liter of air. If the contents of the balloons weight exactly 1.2922 g, then they will have a neutral buoyancy. Let x be the volume of the oxygen balloon, so 1-x is the volume of the hydrogen balloon. The combined weight of the balloons is 1.429*x + 0.08988*(1-x), setting this equal to 1.2922, we get x = 0.8978. so the ratio of the volume of oxygen to hydrogen would be 8.78 : 1. At this ratio, the balloons would be neutrally buoyant and would not rise or sink in the atmosphere (at least not due to buoyancy effects). If the ratio of volumes is increased, they would sink.Note that as temperature and pressure change, the densities will too, and not together either, so neutral buoyancy volume ratio will vary.
CALCULATE THE FINAL VOLUME OF THE BALLOON. HELP ME PLEASE!?
All you need is the following equation: P1 V1 / T1 = P2 V2 / T2, where P1 = init. pressure @ 1.2 atm P2 = final pressure @ 3.00 x 10-3 atm T1 = init. temperature @ 25°C T2 = final temperature @ -23°C V1 = init. volume @ 1.50 L V2 = final volume You want V2, so solve the equation for V2. V2 = (P1 V1 / P2) * (T2 / T1) Then calculate! However, remember the units. Temperature must be in Kelvins. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * If you're wondering where P1 V1 / T1 = P2 V2 / T2 comes from... Start with the Ideal Gas Law. PV = nRT Presuming that there is no leaking as the balloon rises, nR = PV / T n = # moles of gas (should be constant, no leaking) R = the Ideal Gas constant (L * atm / K mol) So the ratio of PV/T should be the same from initial cond. to end because n and R are constants. n*R = P1 * V1 / T1 n*R = P2 * V2 / T2 Therefore, P1 * V1 / T1 = P2 * V2 / T2
Assuming that the balloon is filled with the normal mixture of air and that the skin of the balloon does not compress the air within (as an elastic balloon skin would) The balloon would never rise above the ground. The best gases to fill the balloon to ensure greatest buoyancy would be hydrogen gas. How high would the valloon float? It really depends, because once your balloon floats up to the rarefied air of the upper atmosphere it will expand, if your balloon can expand infinitely, your weightless, absolutely elastic balloon filled with hydrogen gas can float out into space beyond the atmosphere. Realistically speaking, weather balloons filled with hydrogen gases can reach the altitude of around 40km (from wiki)