1. A sample of neon gas at a pressure of 0.687 atm and a temperature of 245 ° C,
ID: 897042 • Letter: 1
Question
1. A sample of neon gas at a pressure of 0.687 atm and a temperature of 245 °C, occupies a volume of 530 mL. If the gas is heated at constant pressure until its volume is 655 mL, the temperature of the gas sample will be _______ °C.
2. A sample of xenon gas at a pressure of 1.16 atm and a temperature of 272 °C, occupies a volume of 470 mL. If the gas is heated at constant pressure until its volume is 741 mL, the temperature of the gas sample will be ________ °C.
3.
Which of the following gas samples would have the largest volume, if all samples are at the same temperature and pressure?
a. 4×1023 molecules of H2
b. 83.8 grams of Kr
c. 0.203 moles CH4
d. they would all have the same volume
4.
Which of the following gas samples would have the largest volume, if all samples are at the same temperature and pressure?
a. 65.43 grams of Xe
b. 3×1023 molecules of O2
c. 0.498 moles CH4
d. they would all have the same volume
Explanation / Answer
1.)
T1 = 245 + 273 = 518 K
V1 = 530 ml
V2 = 655 ml
T2 = ?
formula : V1 /T1 = V2 /T2
530 / 518 = 655 / T2
T2 = 640 .2 K
temperature = 367oC
2)
T1 = 272 + 273 = 545 K
V1 = 470 ml
V2 = 741 ml
T2 = ?
formula : V1 /T1 = V2 /T2
470 / 545 = 741 / T2
T2 = 859 .2 K
temperature = 586oC
3)
3.
answer : b)
a. 4×1023 molecules of H2
1 mol contain ----------> 6.023 x 10^23 molecules
x mol -------------------> 4×1023 molecules
moles of H2 = 4 x 10^23 /6.023 x 10^23 = 0.664 moles
b. 83.8 grams of Kr
Kr moles = mass / molar mass = 83.8/ 83.8 = 1 mol
c. 0.203 moles CH4
according to avagadro law volume is directly propotional to volume. more number of moles more volume
so answer is Kr option b)
4)
.
answer : d. they would all have the same volume
a. 65.43 grams of Xe
Xe moles = 65.43 / 131.29 = 0.498
b. 3×1023 molecules of O2
O2 moles = 3 x10^23 / 6.023 x 10^23 = 0.498 moles
c. 0.498 moles CH4
eqaul moles in all so volume also same
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