Question

A sample of helium has an initial
temperature of 78° C. When the
temperature is lowered to 18,0° C, the
volume of helium becomes 285 mL,
What was the initial volume of the
helium in mL?

Answer 1

Answer:

3460 mL

Explanation:

We are given:

T(i) = 351 k                T(f) = 289 k

V (i) = v mL                V(f) = 285 mL

Using the gas equation:

PV = nRT

Since Pressure (P) , number of moles (n) and R is the universal gas constant

We can say that V/T = k (where k is a constant)

Hence, we can say that:

V(i) / T(i) = V(f) / T(f)

Replacing the values

v / 351  = 285 / 289

v = 3460 mL

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