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Mrac [35]
3 years ago
12

An 8.0 L sample of neon gas at 300. K exerts a pressure of 900. kPa. If the gas is compressed to 2.0 L and the temperature is ra

ised to 600. K, what will be the new pressure?
Chemistry
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:

7200 kPa

Explanation:

Applying,

PV/T = P'V'/T'................ Equation 1

Where P = Initial pressure of neon gas, V = Initial volume of neon gas, T = Initial temperature of neon gas, P' = Final pressure of neon gas, V' = Final volume of neon gas, T' = Final Temperature of neon gas

Make P' the subject of the equation

P' = PVT'/V'T.............. Equation 2

Given: P = 900 kPa, V = 8.0 L, T = 300 K, V' = 2.0 L, T' = 600 K

Substitute these values into equation 2

P' = (900×8×600)/(2×300)

P' = 7200 kPa

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3 years ago
Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm
pav-90 [236]

Answer: 0.0014 atm

Explanation:

Given that,

Original pressure of air (P1) = 1.08 atm

Original volume of air (T1) = 145mL

[Convert 145mL to liters

If 1000mL = 1l

145mL = 145/1000 = 0.145L]

New volume of air (V2) = 111L

New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P2V2

1.08 atm x 0.145L = P2 x 111L

0.1566 atm•L = 111L•P2

Divide both sides by 111L

0.1566 atm•L/111L = 111L•P2/111L

0.0014 atm = P2

Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

7 0
3 years ago
The density of methanol is 0.7918g/mL. Calculate the mass of 89.9mL of this liquid.
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