Answer:
207.03°C
Explanation:
The following data were obtained from the question:
V1 (initial volume) = 6.80 L
T1 (initial temperature) = 52.0°C = 52 + 273 = 325K
P1 (initial pressure) = 1.05 atm
V2 (final volume) = 7.87 L
P2 (final pressure) = 1.34 atm
T2(final temperature) =?
Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:
P1V1/T1 = P2V2/T2
1.05 x 6.8/325 = 1.34 x 7.87/T2
Cross multiply to express in linear form as shown below:
1.05 x 6.8 x T2 = 325 x 1.34 x 7.87
Divide both side by 1.05 x 6.8
T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)
T2 = 480.03K
Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:
°C = K - 273
°C = 480.03 - 273
°C = 207.03°C
Therefore, the final temperature of the gas will be 207.03°C
