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I am doing my chemistry homework and need help with this question. Can someone please help me out and show me how to work this problem out.

A sample of nitrogen has a volume of 7.53 L at 1.715 mmHg and
169 degrees C. What volume will it occupy at standard temperature and pressure?

2006-07-25 13:05:20 · 5 answers · asked by blynn 1 in Science & Mathematics Chemistry

5 answers

Use the ideal gas equation because the no. of moles of nitrogen remains constant i.e. PV/T = nR. At STP, P= 76 mm Hg and T = 273K.
Convert the given temp into Kelvin scale, 169+273=442K

Now, 7.53*1.715/442 = 76*V/273, where V is the volume at STP. Solve for V, which comes out to be ________
Punch the numbers into your calculator

2006-07-25 16:33:12 · answer #1 · answered by skeptical_chymist 2 · 1 1

With Boyle's, Charles, and GayLusaac's Law there are only two variables that are allowed to change. The other two variables were held fixed or constant. This is rather unrealistic since in most cases a sample of gas will be under the influence of all three of the other variables changing. When this happens we are dealing with the Combined Gas Law. The math expression for the Combined Gas Law is as follows:


P1 V1 / T1 = P2 V2 / T2 ; solving for V2=P1V1T2 divided by T1P2
Where P1, T1, and V1 are the original pressure, temperature and volume. T2 = Standard temperature which is O degrees Celsius, and P2 - Standard pressure which is 760mm Hg. Convert your temperatures to Kelvin by adding 273 to each before solving the problem.

2006-07-25 20:50:41 · answer #2 · answered by Shawna 3 · 0 0

P1 x V1/T1 = P2 x V2/T2

Move variables around

V2= P1V1T2/T1P2

Convert Celsius to Kelvin by adding 273 to temp.

V2=(1.715) (7.53) (273)/(444) (760)

and solve

2006-07-25 20:21:00 · answer #3 · answered by I-Eat-Paint-Chips 3 · 0 0

Read the chapter on gas laws. Turn off the TV and the PC and concentrate. Concentrate ...........

2006-07-25 21:03:55 · answer #4 · answered by Anonymous · 0 0

combined gas law

2006-07-26 07:47:48 · answer #5 · answered by x overmyhead 2 · 0 0

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