Question

The melting point of ice at 1 atm is 0 degree celsius. Its melting point will be - 1 degree celsius when pressure of surrounding is:
a. 1 atm
b. 133 atm
c. 272 atm
d. 0.5 atm

Answer 1

Answer:

C. 1 atm (0.99 atm approximately)

Explanation:

according to charles law,

$p1 \div p2 = t1 \div t2 \\ p2 = t1 \times p1 \div t2 \\ p2 = 273k \times 1atm \div 274k \\ p2 = 273atm \div 274 \\ p2 = 0.99atm \: (approx)$

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Answer 2

The pressure of surroundings is 0.996 atm ≅ 1 atm

The initial pressure (P₁) = 1 atm

Initial melting point (T₁) = 0° C

                                     = 273 K

Suppose, the final pressure is = P₂

The final melting point (T₂) = -1° C

                                             = 272 K

Now, if the other properties like, volume etc. remain constant, then,

P ∝ T

∴ P₂/ P₁ = T₂/ T₁

⇒ P₂ = $\frac{272}{273}$ × 1 atm

⇒ P₂ = 0.996 atm

⇒ P₂ ≅ 1 atm

∴ The pressure of the surroundings will near about 1 atm.