Answer:
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
Answer:
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
