An open-too box is formed by cutting equal-sized squares of side length x from each corner of a piece of aluminum measuring 8 inches by 10 inches, then folding up the sides of the box. if the volume needs to be greater than 35 cubic inches, which graph could be used to find possible values of x?

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)

Step-by-step 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;

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