i only have one hour left ! i need help asap

Question

1) A 728 mL gas sample at STP is compressed to a volume of 219 mL, and the temperature is increased to 30◦C. What is the new pressure of the gas in Pa? Answer in units of Pa.

2)A sample of oxygen is collected in a 268 mL container over water at 15◦C, and the barometer reads 756 torr. What volume would the dry gas occupy at 787 torr and 15◦C? Water’s partial pressure at 15◦C is 12.8 torr. Answer in units of mL.

3)Before a trip from New York to Boston, the pressure in an automobile tire is 1.9 atm at 284 K. At the end of the trip, the pressure gauge reads 1.89 atm. What is the new Celsius temperature of the air inside the tire? (Assume tires with constant volume.) Answer in units of ◦C.

4)A helium balloon has a volume of 18.2 L when the pressure is 1.176 atm and the temperature is 19 ◦C. The balloon is cooled at a constant pressure until the temperature is −12◦C. What is the volume of the balloon at this stage?

5)A gas occupying 44 mL at standard conditions is heated to 18◦C while the pressure is reduced to 0.85737 atm. What is the new volume occupied by the gas? Answer in units of mL.

Answer 1

Hello, Leya,

I know my response doesn't arrive within your requested timeframe, but I'd like to share my approach to these questions. Al of these questions deal with a gas that does not change in the number of moles between the two states. That means we can use a simplified version of the gas law:

P1V1/T1 = P2V2/T2

(This combines the ideal gas law equations for the two states, but since the moles, n, and the gas constant, R, don't change, they cancel out.

The temperatures must all be changed to Kelvin for gas law calculations, so add 273.15 to the C value. Then rearrange the equation to isolate the unknown on the left side, and then carefully enter the data. Make sure the units cancel to leave just the one desired. In answering these problems, I'll start with the rearranged equation for that problem.

1. P2 = P1(V1/V2)(T2/T1). Note how I've arranged the volume and temperature conditions into ratios. We can see that the final pressure is the initial pressure times the ratios of the volume and temperature conditions. Directly with temperature (T2/T1) and inversely with volume (V1/V2). Carefully enter the data and solve, We can see all units cancel except temperature, K. The requested pressure is Pascal, so let's use standard pressure of 101.3 kPa for P1. Standard temperature is 0^oC, which needs to be convertrd to Kelvin by adding 273.15. The final temperature is 303.15K.

P2 = P1(V1/V2)(T2/T1)

P2 = (101.3kPa)*(728ml/219ml)*(303.15K/273.15K)

P2 = 373.7 kPa, or 373700 Pa (374 kPa with 3 sig figs)

2. V2 = V1(T2/T1)(P1/P2)

In this problem we need to subtract the partial pressure of the water at 15C, and use the adjusted value for P1. Temperatures are the same, so they cancel. The final volume is reduced since the pressure goes up. I get 253 ml.

3. T2 = T1(V2/V1)(P2/P1)

The initial and final volumes are the same, so that term cancels to 1. We are left only with the pressure change. I get 282.5K

4. V2 = V1(T2/T1)(P1/P2)

I get 11.4 liters for V2

5. V2 = V1(T2/T1)(P1/P2)

I get a new volume of 49.1 ml. The change in temperature has a lesser impact than the change in pressure. (Consider the relative value of those two ratios. T2/T1 is 0.956, and P1/P2 is 1.17)

I hope this helps. Look at the ratios of the rearrnaged equations to make a judgement on the direction of the change (up or down) and also get a feel for how much that change will be.

Bob