Chemistry, 31.01.2021 08:30 jordanmartinezacierr

8) A gas that has a volume of 28 liters, a temperature of 65 0C, and an unknown pressure has its volume increased to 36 liters and its temperature decreased to 35 0C. If I measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas?

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Answer from: deadea46

The original pressure of the gas whose pressure changed to 2 atm is 2.82atm.

HOW TO CALCULATE PRESSURE:

The pressure of a gas can be calculated by using the following expression:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressureP2 = final pressureV1 = initial volumeV2 = final volumeT1 = initial temperatureT2 = final temperatureAccording to this question; P1 = ?, P2 = 2 atm, V1 = 28L, V2 = 36L, T1 = 65°C (338K), T2 = 35°C (308K)P1 × 28/338 = 2 × 36/3088624P1 = 24336P1 = 24336 ÷ 8624P1 = 2.82atmTherefore, the original pressure of the gas whose pressure changed to 2 atm is 2.82atm.

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Answer from: tanviknawale

P₁ = 2.82 atm

Explanation:

Given data:

Initial volume = 28 L

Initial pressure = ?

Initial temperature = 65 °C (65 +273.15 = 338.15 K)

Final temperature = 35 °C (35 +273.15 = 308.15 K)

Final volume = 36 L

Final pressure = 2.0 atm

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂V₂T₁/T₂ V₁

P₁ = 2.0 atm × 36 L × 338.15 K / 308.15 K × 28 L

P₁ = 24346.8 atm .L. K / 8628.2 K.L

P₁ = 2.82 atm

Answer from: Quest

possibly, yes.

explanation:

the amount of sugar you take in is dangerously high, risk of heart disease, high risk of diabetes, and etc.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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8) A gas that has a volume of 28 liters, a temperature of 65 0C, and an unknown pressure has its vol...

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