13. a gas occupies 4.31 liters at a pressure of 0.755 atm. determine the volume if the pressure is increased to 1.25 atm. 14.

3 answers
Question:

13. a gas occupies 4.31 liters at a pressure of 0.755 atm. determine the volume if the
pressure
is increased to 1.25 atm.
14. 600.0 ml of a gas is at a pressure of 8.00 atm. what is the volume of the gas at 2.00
t
.

Answers

13. 2.60 L.

14. 2.40 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and T are constant, and have different values of P and V:

(P₁V₁) = (P₂V₂)

13. A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the  pressure  is increased to 1.25 atm.

P₁ = 0.755 atm, V₁ = 4.31 L.

P₂= 1.25 atm, V₂ = ??? L.

∴ V₂ = (P₁V₁)/(P₂) = (0.755 atm)(4.31 L)/(1.25 atm) = 2.60 L.

14. 600.0 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at 2.00  at m.

P₁ = 8.0 atm, V₁ = 600.0 mL.

P₂= 2.0 atm, V₂ = ??? L.

∴ V₂ = (P₁V₁)/(P₂) = (8.0 atm)(600.0 mL)/(2.0 atm) = 2400/0 mL = 2.40 L.

1. A gas occupies 4.31 litres at a pressure of 0.755atm. Determine the volume if the pressure is increased to 1.25 atm.

method: Boyles Law: P1/V1 = P2/V2
P1/V1 = P2/V2 implies V2= P2 V1/ P1
V2=1.25x4.31 / 17=7.13L


2. 600 mL of a gas is at a pressure of 8.00atm. What is the Volume of the gas at 2 atm?

method: Boyles Law: P1/V1 = P2/V2
P1/V1 = P2/V2 implies V2= P2 V1/ P1,
V1=2.6mL=0.026L
V2= P2 V1/ P1= 0.00065 L

3. How hot will a 2.3 L balloon have to get to expand to a volume of 400 L? Assume that the initial temperature of the balloon is 25 degrees Celsius.

Method: Charles Law: V1/T1 =V2/T2  Degrees Celcius + 273 = Kelvin

25 degrees Celsius=25+273=298K=T1

V1=2.3L, V2=400L

V1/T1 =V2/T2 implies T2=V2T1/V1=(400x298) /2.3=51826 K

4. A gas takes up a volume of 17 litres, has a pressure of 2.3atm, and a temperature of 299K. If I raise the temperature to 350K and lower the pressure to 1.5atm, what is the new volume of the gas?

Combined gas Law: (P1)(V1) =  (P2)(V2)
                                     (T1)           (T2)

V1=17L
P1=2.3 atm
T1= 299K
P2=1.5 atm
T2=350 K
V2=?

 (P1)(V1) / T1=  (P2)(V2)/ T2
                                 
so V2= (P1)(V1) T2 / P2 T1
V2=17x2.3x350 /1.5 x 299= 30.51L

The new volume at standard temperature and pressure is 5.08 L.

Explanation:

As per the kinetic theory of gases, the volume occupied by gas molecules will be inversely proportional to the pressure of the gas molecules. This is termed as Boyle's law.

So, pressure∝[tex]\frac{1}{volume}[/tex]

Thus, if two pressure and two volumes are given then,

[tex]P_{1} V_{1} = P_{2} V_{2}\\[/tex]

Now, we known the values of P₁ = 8 atm, V₁ = 635 mL, P₂ = 1 atm and V₂ we have to determine. We are considering P₂ = 1 atm, because we have to determine V₂ at standard temperature and pressure. And standard pressure is 1 atm.

[tex]8*635*10^{-3} = 1 *V{2} \\\\V_{2} = \frac{8*635*10^{-3} }{1} = 5.08 L[/tex]

Thus, the new volume at standard temperature and pressure is 5.08 L.

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