A quantity of gas occupies a volume of 506 cm 3 at a temperature of 147 o C.

Assuming the pressure stays constant, at what temperature will the

volume of the gas be 604 cm 3?

228.3°C

Explanation:

Data obtained from the question include:

V1 (initial volume) = 506 cm3

T1 (initial temperature) = 147°C = 247 + 273 = 420K

V2 (final volume) = 604 cm3

T2 (final temperature) = ?

The gas is simply obeying Charles' law because the pressure is constant.

The final temperature of the gas can be obtained by using the Charles' law equation V1/T1 = V2/T2 This is illustrated below:

V1/T1 = V2/T2

506/420 = 604/T2

Cross multiply to express in linear form as shown:

506 x T2 = 420 x 604

Divide both side by 506

T2 = (420 x 604) / 506

T2 = 501.3K

Now let us convert 501.3K to a temperature in celsius scale. This is illustrated below:

°C = K - 273

°C = 501.3 - 273

°C = 228.3°C

Therefore, the temperature of the gas when the volume of the gas is 604 cm3 is 228.3°C

180°C

Explanation:

Here we can apply Charles' law which states that the volume of a fixed mass of gas is directly proportional to its temperature at constant pressure

Symbolically written as

VαT

V=KT

V/T = K

V1/T1 = V2/T2

We shall apply this formula derived from Charles' law to this question since pressure is kept constant

Step 1

From the question,

V1 = 506cm³

V2 = 604cm³

T1 = 147°c = 147 + 273 = 420K

T2 = ?

Recall, V1/T1 = V2/T2

506cm³/420K = 604cm³/T2

We make T2 the subject by cross multiplying, then dividing

506cm³*T2 = 604cm³*420K

T2 = (604cm³*420K) / 506cm³

T2 = 501.3K

501.3K = 501.3 - 273 = 228.3°C

Therefore, the temperature of the gas at a volume of 604cm³ is 501.3K or 228.3°C