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PLEASE HELP ME FAST, PLEASE

PLEASE HELP ME FAST, PLEASE-example-1

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Answer:

The temperature order are;

(b) 96.62 K = (d) 96.62 K > (a) 48.31 K = (c) 48.31 K = (e) 48.31 K = (f) 48.31 K

Arrangement in order from highest to lowest and alphabetically gives;

(b) ↔ (d) → (a)↔ (c)↔ (e)↔ (f)

Explanation:

From the universal gas equation

P×V = N×k×T

Where:

P = Pressure

V = Volume

N = Number of molecules

k = Boltzmann constant = 1.38 × 10⁻²³ J/K

T = temperature

Therefore;


T =(P * V)/(k * N)

Which gives;

(a) When P = 100 kPa = 100,000 Pa, V = 4 L = 0.004 m³, N = 6 × 10²³, we have

100000*0.004/(6*10^(23)*1.38*10^(-23)) = 48.31 K

(b) When P = 200 kPa = 200,000 Pa, V = 4 L = 0.004 m³, N = 6 × 10²³, we have

200000*0.004/(6*10^(23)*1.38*10^(-23)) = 96.62 K

(c) When P = 50 kPa = 50,000 Pa, V = 8 L = 0.008 m³, N = 6 × 10²³, we have

50000*0.008/(6*10^(23)*1.38*10^(-23)) = 48.31 K

(d) When P = 100 kPa = 100,000 Pa, V = 4 L = 0.004 m³, N = 3 × 10²³, we have

100000*0.004/(3*10^(23)*1.38*10^(-23)) = 96.62 K

(e) When P = 100 kPa = 100,000 Pa, V = 2 L = 0.002 m³, N = 3 × 10²³, we have

100000*0.002/(3*10^(23)*1.38*10^(-23)) = 48.31 K

(f) When P = 50 kPa = 50,000 Pa, V = 4 L = 0.004 m³, N = 3 × 10²³, we have

50000*0.004/(3*10^(23)*1.38*10^(-23)) = 48.31 K

User Cos Callis
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