D. speed E. size molecules 14, Agas sample with a volume of 4.0 L has a pressure

Question

D. speed E. size molecules 14, Agas sample with a volume of 4.0 L has a pressure of 750 mmHg and a temperature of 77 °C. what is its new volume at 277 and 250 mmHg (n constant) ? (85) B. 19LDP C. 2.1 L A. 7.6 L D. 0.00056 L E. 3.3 LI T 15. If the temperature and amount of a gas does not change, but its volume doubles, its pressure will (8.2) 60 X A. double B. triple C. decrease to one-half the original pressure D. decrease to one-fourth the original pressure E. not change 16. A tank of 02 used at home for a patient with pulmonary disease has a pressure of 220 mmHg, 2.0 moles of gas, and a volume of 4.0 L. What will the new pressure be when the volume expands to 5.0 L and 3.0 moles of O2 is added, if temperature is constant? (8.6) A. 110 mmHg D. 550 mmHg C. 440 mmHg B. 220 mmHg E. 700. mmHg 17. A sample of 2.00 moles of gas initially at STP is converted to a volume of 5.0 L and a temperature 2 of 27 °C. What is its new pressure in atmospheres? (8.5, 8.6) C. 7.5 atm A. 0.12 atm D. 9.8 atm The conditions for standard temperature and pressure (STP) are (8.6) B. 5.5 atnm E. 10. atm A. 0K, I atm D. 273 K, 1 atm C. 25 °C, 1 atm B. 0°C, 10 atm E. 273 K, 0.5 atm 19. The volume occupied by 1.50 moles of CH, at STP is (8.6) 2 B. 33.6L E. 5.60 L A. 44.10 L C. 22.4L How many grams of O, are present in 44.1 L of O, at STP? (8.6)

Explanation / Answer

16.) PV = nRT (ideal gas equation)

=> PV/ n = RT  

Since R and T are constant, means right side is constant. So left side will be constant throughout the process.

P1 V1 / n1 = P2 V2 / n2

P1 = 220 , V1 = 4 , n1 = 2

P2 = ? , V1 = 5, n2 = 2+3 = 5

Put it in the above equation and calculate P2

so, 220 x 4 / 2 = P2 x 5 / 5

P2 = 440 mm Hg

17) STP = 1 atmosphere pressure and 273 Kelvin temperature (0°C).
1 mole at STP = 22.4 Litres Volume.
2 moles therefore = 44.8 Litres

Same formula but here moles are constant. so, PV/T = constant

1 x 44.8 / 273 = P2 x 5 / 300 [convert 27 celcius to kelvin which is 300K]

so, P2 = 9.84 atm ~ 10 atm

so answer is E.

19) again at STP = 1 atmosphere pressure and 273 Kelvin temperature (0°C).
1 mole at STP = 22.4 Litres Volume.
1.5 moles therefore = 22.4 x 1.5 = 33.6 Litres

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