5. You have a 965 mL sample of methane at 92°C and a pressure of 1.2910, tom. A.

Question

5. You have a 965 mL sample of methane at 92°C and a pressure of 1.2910, tom. A. If you lower the temperature to -43°C without changing the volume, what pressure will the methane exert? B. Using the results from part A, if you then decrease the volume to 623 mL while keeping the temperature at -43°C, what pressure will the methane exert? C. Using the results from part B, if you now increase the temperature to 83'C while keeping the pressure constant, what will the volume of the methane be? D. If you add 1.2 moles of oxygen to the methane in part C while keeping temature and volume constant, what will be the partial pressure of methane, the partial pressure of oxygen, and the total pressure? E. What is the mass of the oxygen?

Explanation / Answer

Now from ideal gas equation we know that PV= nRT

P is pressure, v is volume , n is number of moles, T is temp and and R is the gas constant. Here T is in kelvin

If the volume remains constant T1/P1=T2/P2

Thus (92+273)/(1.29*10^3)=(-43+273)/P2

Thus P2 is 0.81x10^3 torr

Now if the temp remains constant we can use P3V3=P2V2

thus putting values P3x623=0.81x10^3x965 we get P3=1.26x10^3 torr

Now P is constant so we use T3/V3=T4/V4

Thus (-43+273)/623= (83+273)/V4

V4=964.29 ml

If temp and volume are the same then partial pressure exerted is directly proportional to the number of moles of the gas.

Calculating pressure exerted by O2 using ideal gas equation

Px0.964= 1.2x0.0821x356

P= 36.38 atm = 760x36.38 torr =27.6x10^3 torr

Total pressure is (1.26+27.6)x10^3 torr = 28.86 x10^3 torr

Mass of oxygen = moles*molar mass = 1.2x 32 = 38.4 gms

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