A certain amount of a perfect gas occupies a volume of 15.0 dm\' at a temperatur

Question

A certain amount of a perfect gas occupies a volume of 15.0 dm' at a temperature of 1.1 (25) 240 K and at a pressure of p -420 kPa. Ir this gas is compressed into a volume of 3150 cm at this temperature, (a) to what value will the pressure be changed. Determine the number of moles in this sample at room temperature. 1.2025s) Use the latest weather report (newspaper or website) to determine the local ambient temperature and pressure. Convert these data to units of K and Pa, respectively. For these conditions, determine the volume that I mol of a perfect gas will occupy. State your answer in units of m', dm' and cm' 1.3 G5v) At a constant temperature, a perfect gas is being compressed and as a result its volume decreases by 1.8 dn'. The final pressure and volume of the gas are 142 Torr and 6.12 L respectively. Calculate the original pressure of the gas in (a)Torr,(b bar. (c) Pa, (d) atm. 1.4 (15%) To what temperature must a sample of a perfect gas of volume 600 ml be cooled from 45C in order to reduce its volume to 120 cm' under isobaric conditions? Provide the answer in oC and in K units. mportant note: PRINT your name and Posting-ID number in the top right comer of the first page. No electronic ubmissions. Please write clearly, show your work including all units, and state your answer in terms of gnificant digits. To simplify grading, please mark the final answer of each problem with a box around it.

Explanation / Answer

1.1:(a): Initial condition:

Given Volume, V1 = 15.0 dm3 = 15.0 dm3 x (1L / 1 dm3) = 15.0 L

Temperature, T = 240 K

Pressure, P1 = 420 KPa = 420 KPa x (0.00986923 atm / 1 KPa) = 4.145 atm

Final condition: Volume, V2 = 3150 cm3 = 3150 cm3 x (1L / 1000 cm3) = 3.15 L

Let the final pressure be P2 atm.

Given the temperature is constant.

Applying Boyle' s law

P1V1 = P2V2

=> P2 = P1V1 / V2 =  420 KPa x 15.0 L / 3.15 L = 2000 KPa = 19.74 atm (answer)

(b): Room temperature, T1 = 298 K

Applying ideal gas equation

P1V1 = nRT1

=> n = P1V1 / RT1 = (4.145 atm x 15.0L) / (0.0821 L.atm.mol-1K-1 x 298K) = 2.54 mol (answer)

1.3: (a): Let original pressure be 'P1' atm

Given final pressure, P2 = 142 torr

Final volume, V2 = 6.12 L

Given volume is increased by 1.8 dm3. Hence

V2 - V1 = 1.8 dm3 = 1.8 dm3 x (1L / 1dm3) = 1.8 L

=> 6.12 L - V1 = 1.8 L

=> V1 = 4.32 L

Now applying Boyle's law for the above condition

P1V1 = P2V2 (T = constant)

=> P1 = P2V2 / V1 = (142 torr x 6.12L) / 4.32L = 201 torr (answer)

(b)  P1 =  201 torr =  201 torr x (0.00133322 bar / 1 torr) = 0.268 bar (answer)

(c) P1 =  201 torr = 201 torr x (133.322 Pa / 1 torr) = 26798 Pa (answer)

(d): P1 = 201 torr = 201 torr x (0.00131579 atm / 1 torr) = 0.264 atm (answer)

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