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Assoli18 [71]
4 years ago
6

A sample of the inhalation anesthetic gas Halothane, in a 500-mL cylinder has a pressure of 2.3 atm at 0°C. What will be the pre

ssure of the gas if its temperature is warmed to 37°C (body temperature)? (Hint: Which law will you apply if the volume is constant?)
Chemistry
2 answers:
STALIN [3.7K]4 years ago
8 0

Answer: 2.56atm

Explanation:

P1/T1=P2/T2

P1T2=P2T1

P2=P1T2/T1

P1=2.3atm

P2=?

T1=0+273=273k

T2=32+273=305k

P2=2.3*305/273

P2=701.5/273

P2=2.56atm

lilavasa [31]4 years ago
3 0

Answer:

The pressure will increase to a pressure of 2.6 atm

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500L

Pressure = 2.3 atm

Temperature = 0°C = 273 K

Temperature is warmed to 37 °C = 310 K

Volume will stay constant

Step 2: Calculate the new pressure

p1/T1 = p2/T2

⇒with p1 = the initial pressure = 2.3 atm

⇒with T1 = the initial temperature = 0 °C = 273 K

⇒with p2 = the new pressure = TO BE DETERMINED

⇒with T2 = the new temperature = 37 °C = 310 K

2.3 atm / 273 K  = p2 / 310 K

p2 = 2.6 atm

The pressure will increase to a pressure of 2.6 atm

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An enclosed vessel contains 2.5g of 9b nitrogen and 13.3g of chlorine at s.T.P. Of What will be the partial pressure of the Il n
kow [346]

Answer:

0.535 atm

Explanation:

Since the volume of the tank is constant, we use Gay- Lussac's law to find the pressure at 180°C.

So, P₁/T₁ = P₂/T₂ where P₁ = pressure at S.T.P = 1 atm, T₁ = temperature at S.T.P = 273.15 K, P₂ = pressure of gas at 180 °C and T₂ = 180 °C = 273.15 + 180 K = 453.15 K

So, P₁/T₁ = P₂/T₂

P₂ = P₁T₂/T₁

Substituting the values of the variables into the equation, we have

P₂ = P₁T₂/T₁

P₂ = 1 atm × 453.15 K/273.15 K

P₂ = 1 atm × 1.66

P₂ = 1.66 atm

We now need to find the total number of moles of each gas present

number of moles of nitrogen = mass of nitrogen, m/molar mass of nitrogen molecule M

n = m/M

m = 2.5 g and M = 2 × atomic mass of nitrogen (since it is diatomic) = 2 × 14 g/mol = 28 g/mol

So, n = 2.5 g/28 g/mol

n = 0.089 mol

number of moles of chlorine, n' = mass of chlorine, m'/molar mass of chlorine molecule M'

n' = m'/M'

m' = 13.3 g and M = 2 × atomic mass of chlorine (since it is diatomic) = 2 × 35.5 g/mol = 71 g/mol

So, n' = 13.3 g/71 g/mol

n' = 0.187 mol

So, the total number of moles of gas present is n" = n + n' = 0.089 mol + 0.187 mol = 0.276 mol

So, the partial pressure due to nitrogen gas, P = mole fraction of nitrogen × pressure of gas at 180 °C

P = n/n" × P₂

P = 0.089 mol/0.276 mol × 1.66 atm

P = 0.322 × 1.66 atm

P = 0.535 atm

8 0
3 years ago
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