"A gas occupies a volume of 400mL at 500 mmHg pressure. What will be the volume if changed to 760 torr?
476691
As long as you are not changing the temperature or the number of moles of gas in the system, the following equation applies:\n\nP1V1 = P2V2\n\nthus (400 mL)*(500 torr) = (X mL)*(760 torr)\n\nSolve for X.
765230
EER = Q in (BTU/hr) divided by W in (watts)\n\n600/500= 1.2
1097483
P/T=P/T\n1/(20+273)=P/(600+273)\nP=30 atm\n\nThe new pressure would be 30 atms, but this is considering that the volume remains constant.
658673
As someone said, this is a mass not a volume but that doesn't mean you can't do it : this is all about conversions.\n\n1) Gas : The only thing you need to know is that a gram mole of any gas at standard conditions occupies 22.41 liters. You can work out the equivalent volumen for a kilogram mole for yourself (hint, 1000 time as much) and don't forget to change the molar volume (use the ideal gas law) for the conditions if they are not standard. This will be to a first approximation the same volume for a kmol of any gas (providing its not at not too extreme conditions)\n\n2) liquid : you need to work out the molar volume from the density and the molecular weigh. If you don't have that data, it can't be done. Unlike the gas, this cam be very differnent for different liquids, and also chages with conditions but in a manner that cannot be calculated with a simple formula \n\nWhe you come to do real work with any of these things, you usually find that volumes are given in standard and actual conditions. Be careful to use the conversion factors at the appropraite conditions or you will get it wrong\n\nEnjoy.
279834
500 mm Hg is 500 torr, check http://www.lds-vacuum.com/conversion.html if your not sure.
683503
In the relationship between pressure, temperature and volume of gases, the pressure and temperature have a direct relationship....if one factor goes up so does the other one.  So, set up what you know as a ratio.\n\npressure/temperature = pressure/temperature\n\n400/80  =   300/x       cross multiply and solve for unknown\n\n24000 = 400x\n\n60 = x
332490
320 gm = x mL * 0.63 g/mL\nx = 320/0.63 = 32000/63 gm = 507.936 gm
170709
Use the ideal gas law: PV=nRT where P is pressure in Pa, V is volume in meters cubed (m^3), n is amount in moles, T is temperature in Kelvin (K), and R is the ideal gas constant 8.314 J/mol K.\n\nSolving for V: V=nRT/P\nYou have to convert pressure from atm to Pa. The conversion factor is 1 atm = 101325 Pa\n\nV=(2.44 mol)(8.314 J/mol K)(343 K)/243180 Pa\n\nV=0.0286 m^3
700450
Letters A, B, D.\nIn not sure about C, its wording is tricky.\nIdeal gas law : PV = nRT\nReal gas law : (P + (a(n / V)^2)) * (v - nb) = nRT\nThe corrected pressure times the corrected volume.
777799
900 torr = 1.184 atm\nBoyle's Law:\nP1V1=P2V2\n\n(1.184atm) (V1) = (1.50atm) (.25 L)\n\nV1 = .317 L = 317 mL\n\nYour welcome.
121351
Are you tryn to have us do your hw!?!?\n\nThe eqn should be in your book.  I dont feel like digging mine out, but its something like;\n\nCp*g*(T2-T1)=energy\n\nSo what u want to find is (t2-t1) Ima call it x\n\n400*50*(x)=800  do the math.  I think thats it.\n\nfor the 2nd question, this one is easy.  The combustion is energy is for 1 mol (per mol).  If you only have .2 moles, the combustion amount is .2* 1370 .
355529
V(gas)=0,2 m3,T1=333 K,T2=533 K,V2=?\nidel gas law: pV=nRT,p=const. so nRT1/V1=nRT2/V2\nT1/V1=T2/V2, V2=T2V1/T1\nV2=533 K x 0,2 m3 /333 K =0.32 m3
1140114
What's the question?\n\nOk, 30 / (12*6+6) = 0.385 mol\n\n0.385 X 2/1 = 0.770 mol of C6H5Br, which makes 120.89 g.
129942
the ideal gas law is:\n \n PV = nRT \n\nSO,\n\n P= .95 atm\n n=  .87\n T=  280K\n R=  .0821\n V=  ?\n\n solve for V\n\n     V = nRT/P =(.87* .0821* 280) /.95 = 21 liters
933724
We will assume CH4 is an ideal gas.\n\nPV=nRT\nP=nRT/V= 0.123mol * (0.0821 atm*L/(mol*K)) * 760 torr/1atm * (30+273.15)K / 20.0L = 116 torr\n\nUsing Dalton's Law of partial pressures, Ptotal = P1 + P2 + P3 + .... + Pn, the total vapor pressure is just the sum of the two partial pressures.\n\nP= 31.82 torr + 116 torr = 148 torr
134712
This question is based on the gas law PV=nRT where \n\nP= pressure in atm (1atm=101.3kPa)\nV= volume in litres\nn= # of moles\nR= .08206\nT= temp. in Kelvin (273.15K=1*C)\n\nand the equation\n\nP1V1=P2V2
739100
I can't answer the question, but you need Boyle's Law, the volume of a gas is inversely proportional to it's pressure.  Also, are you assuming there is no temperature change?