A sample of air has a volume of 25 L at 1.2 atm and 37 degree C. If the sample i

Question

A sample of air has a volume of 25 L at 1.2 atm and 37 degree C. If the sample is transformed completely to a container with a volume of 10 L at standard pressure, what will be the new temperature? 103 k 111 k 12.3 k 212 k What is the mass of a sample of Neon gas if the volume is 10 L at STP? 9.0 g 0.04 g 0.45 mol 0.82 mol A gas confined in a 515 cm^2 container exerts a pressure of 107.4 kPa at 38.6 degree C. At what Kelvin temperature will it exert a pressure of 635.7 kPa if it is placed into a 644 cm^2 container? 286 K 2306 K 0.00043 K 5.2 K What volume would be occupied by 100.0 g of oxygen gas at a pressure of 1.50 atm and a temperature of 25 degree C? 51 L 1630 L 137 L 6L How many moles of O_2 are in a sample with a pressure of 101,325 kPa, a volume of 25 L and a temperature of 0 degree C? 113,156 mol 1116.1 mol 34.9 mol 34.9 g A gas is confined in a container with a volume of 25 L at 273 K and 1 atm of pressure. What will be the new pressure if the temperature is changed to 373 K and volume is decreased to 10 L? 1.8 atm 1.0 atm 3.4 atm 0.3 atm A 35.0 L tank contains 20 g of nitrogen gas at standard pressure. What will be the temperature in Kelvin? 21.3 K 6 K 601 K 700 K A rigid container holds ammonia gas at a presure of 55 kPa and a temperature of -100 degree C. What will the pressure be when the temperature is increased to 200 degree C? 20 kPa 63.6 kPA -110 kPa 150.4 kPa

Explanation / Answer

1)a)103K

v1 = 25,p1 =1.2atm,T1 = 310K
V2 = 10L,p2 =1atm,T2 = /
p1v1/T1 = p2v2/T2
1.2*25/310 = 10/T2
T2 = 103.33K

2)a)

x/20 = n
at STP
v = 10L
P = 1atm
T = 273K
22.4*1 = (x/20) *0.082*273
x = 10 *20 /(0.082*273)
x = 8.93 grams

3)b)0.00043K

v1 = 515cm^3
p1 = 107.4Kpa
T1 = 311.6K
v2 = 644cm^3
P2 = 635.7Kpa
T2 = ?
p1v1/t1 = p2v2/t2
515*107.4/311.6 = 635.7*644/T2
T2 = 0.00043K

4)a) 51L

v= ?,n = 100/32 = 3.125
p =1.50 atm,T = 298K
v = nRT/p = 3.125*0.082*298 / 1.50 = 50.90L

5)C)1116.1 mol

1*25 = n(0.082)*273
n = 25/0.082*273

6)b)3.4

v= 25L,T= 273K,p = 1atm
p2=?,T2 = 373k,v2 = 10L
p1v1/T1 = p2v2/T2
1*25/273 = p2*10/373
p2 = 373*25/10*273 = 3.4atm

7)b

pv = nRT

1*35 = 20/28*0.082*T

35 = 0.714*0.082*T

T = 35/ (0.714*0.082)

T = 598K

8) d

p1/T1 = p2/t2

55kPA/173 = p2/473

p2 = 55*473 / 173 = 150.4 Kpa

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