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Elanso [62]
3 years ago
10

Nitrogen gas is introduced into a large deflated plastic bag. No gas is allowed to escape, but as more and more nitrogen is adde

d, the bag inflates to accommodate it. The pressure of the gas within the bag remains at 1.00 atm and its temperature remains at room temperature (20.0∘C). How many moles n have been introduced into the bag by the time its volume reaches 22.4 L
Chemistry
1 answer:
Natalija [7]3 years ago
4 0

Answer:

0.9307 moles have been introduced into the bag.

Explanation:

Pressure of the gas within the bag,P = 1.00 atm

Temperature of the gas remains at room temperature,T=20.0 °C = 293.15 K

Volume of the gas in the bag = V = 22.4 L

Number of moles of gas = n

Using an ideal gas equation:

PV=nRT

1.00 atm\times 22.4 atm=n\times 0.0821 atm l/mol K\times 293.15 K

n = 0.9307 moles

0.9307 moles have been introduced into the bag.

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Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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