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Andrei [34K]
3 years ago
8

A gas has a volume of 350.0 mL at 45.0°C. If the volume changes to 400.0 mL, what is the new

Chemistry
1 answer:
nekit [7.7K]3 years ago
5 0

Answer:

363.6 K

Explanation:

To solve this problem we will use Charles law equation i.e,

V1/T1 = V2/T2    

Given data

V1= 350 mL

T1= 45.0 °C

V2= 400.0 mL

T2=?

First of all we will convert the temperature into kelvin because the kelvin is an absolute scale of temperature.

T1= 45.0+ 273.15 = 318.15 K

Now we will put the values in equation

350 mL / 318 K = 400 mL / V2

V2= 400 mL × 318.15 K / 350 mL  

V2= 363.6 K

We can see that by increasing the temperature volume of given gas is also increases. we conclude that volume and temperature are directly proportional to each other. It is also confirmed by the Charles law.

“ The volume of given mass of a gas at a constant pressure is directly proportional to the absolute temperature ”

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In the gas-phase reaction 2 A + B ⇌ 3 C + 2 D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and all
Naddik [55]

Answer:

(i)The mole fractions are :

  • A=\frac{0.4}{4.3} \\=0.0930
  • B=\frac{1.4}{4.3} \\=0.3256
  • C=\frac{0.9}{4.3} \\=0.2093
  • D=\frac{1.6}{4.3} \\=0.3721

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Explanation:

The given equation is :

2A+B⇄3C+2D

Let \alpha be the number of moles dissociated per mole of B

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2A\\(1-2\alpha)     +  B\\2(1-\alpha)  ⇄  3C\\(3\alpha)   + 2D\\(1+2\alpha)

And finally the number of moles of C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930

  • B=\frac{1.4}{4.3} \\=0.3256
  • C=\frac{0.9}{4.3} \\=0.2093
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(ii)

K=\frac{(P_C^3)(P_D^2)}{(P_A^2)(P_B)}

Where ,

P_A,P_B,P_C,P_D are the partial pressures of A,B,C,D respectively.

Total pressure = 1 bar .

∴

<em>P_A= 0.0930*1=0.0930</em>

<em>P_B= 0.3256*1=0.3256</em>

<em>P_C= 0.2093*1=0.2093</em>

<em>P_D= 0.3721*1=0.3721</em>

K=\frac{0.2093^3*0.3721^2}{0.0930^2*0.3256} \\K=0.4508

(iii)

ΔG=-RTlnK\\

ΔG = -8.314*(273+25)*ln(0.4508)\\=1973.96J\\=1.974kJ

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