True A sample of gas at 200 degree C has a volume of 5 L. The volume expands to

Question

True A sample of gas at 200 degree C has a volume of 5 L. The volume expands to 10 L due to a temperature change at constant pressure. What will be the new temperature? -36.5 degree C 673 degree C 400 degree C 100 degree C 0 degree C Before a trip from New York to Boston, the pressure in an automobile tire is 1.9 atm at 270 K. At the end of the trip, the pressure gauge reads 1.9 atm. Wliat is the new Cel­sius temperature of the air inside the tire? (Assume tires with constant volume.) Answer in units of degree C At 141 degree C, the pressure of a sample of nitrogen is 1.9 atm. What will the pressure be at 257 degree C, assuming constant volume? Answer in units of atm A sample of helium gas has a pressure of 1.4 atm at 1 degree C. At what Celsius temperature will the helium reach a pressure of 2.12 atm? Answer in units of degree C An empty aerosol-spray can at room temper­ature (20 degree C) is thrown into an incinerator where the temperature reaches 427 degree C. If the gas inside the empty container was initially at a pressure of 1 atm, what pressure did it reach inside the incinerator? Assume the gas was at constant volume and the can did not explode. The volume of a gas at 27 degree C and 0.32 atm is 31 mL. What volume will the same gas sample occupy at standard conditions? Answer in units of mL A gas has a volume of 1.46 L at -13 degree C and 163 kPa. At wliat temperature would the gas occupy 1.2 L at 228 kPa? Answer in units of degree C For a certain ideal gas, the temperature is increased from 200 K to 541 K allowing pres­sure, which is initially 1 atm. to vary while the volume and number of moles of gas are held constant. What is the new pressure? Answer in units of atm

Explanation / Answer

12) Temperature is directly proportional to Volume

T1 = 200 + 273 = 473 degrees

T2 shall be double of this temperature = 2 * 473 = 946 K = 946 - 273 = 673 degrees celcius

Option 2

13) Gauge pressure = 1.9 atm

Thus absolute pressure = 1.9 + 1.0 = 2.9 atm = P2

P1 = 1.9 atm

now, P1 / P2 = T1 / T2

Thus, T2 = 270 * 2.9/1.9 = 412.10 K

412.10 K = 139.10 degree celcius

14) 141 degree = 414 K = T1

257 degrees = 530 K = T2

T1/T2 = P1 / P2

P1 = 1.9 atm

P2 = P1 * (T2/T1) = 1.9 * 530/414 = 2.432 atm

15) P1 = 1.4 atm

P2 = 2.12 atm

T1 = 1 + 273 = 274 K

T2 = T1 (P2/P1) = 414.91 K = 141.9 degrees celcius

16) P1 = 1 atm T1 = 20 + 273 = 293 K

T2 = 427 + 273 = 700 K

P2 = P1( T2/T1) = 2.389 atm

17) P1V1 / T1 = P2V2 / T2

0.32 * 34 / 300 = 1 * V1/ 273

V1 = 9.9 ml

18) P1V1 / T1 = P2V2 / T2

1.46 * 163 / 260 = 1.2 * 288 / T2

T2 = 104.57 degrees celcius

19) P1 /P2 = T1 /T2

1 / P2 = 200 / 541

P2 = 541 / 200 = 2.705 K

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