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There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?

User Novalis
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1 Answer

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Answer:

We are given:

V1 = 5.5L T1 = -38 C or 235 k

V2 = 1.3L T2 = T

From the gas equation:

PV = nRT

Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:

V / T = k (where k is a constant)

so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables

Changing the variables for the first case:

V1 / T1 = k (where k is the same constant) ----------------(1)

Similarly,

V2 / T2 = k (again, k has the same value)----------------(2)

From (1) and (2):

k is the common value

V1 / T1 = V2 / T2

Replacing the variables

5.5 / 235 = 1.3 / T

T = 1.3 * 235 / 5.5

T = 55.54 k

Therefore, at 55.54 K the gas will have a volume of 1.3L

User Pramodh Valavala
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