If 30cm\(^3\) of gas at 50\(^0\)C is warmed to 80\(^0\)C at a fixed pressure, the fractional increase in volume is
0.009
0.093
0.910
1.090
Explanation
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No,myschool they made a mistake. The answer is B which is 0.093.
I will explain myself.
Using Charles' Law,
V1/T1=V2/T2
30/323=V2/353.
323 and 353 gotten from 50°C + 273°C=323°C
likewise;
80°C+273°C=353°C.
Therefore;
New volume =32.786cm³
Old volume =30.00cm³
New vol/Old vol= [alpha - 1]
so; 1.093-1 =0.093.
The question seem higher for O'Level students.
Myschool check and correct it
Please the answer is B. Use of Charles law
V1/T1 = V2/T2
30/(50+273) = V2/(80+273)
V2=30*353/323
=32.78
fraction increase= (32.78-30)/30
=0.093
please correct my school
The fractional change of a quantity refers to the ratio of the change in the quantity to its original value. It measures the relative size of the change compared to the original value. The formula for calculating the fractional change is: fractional change = (change in quantity) / (original quantity)
So, it's simple fraction that's causing chaos like this????
Answer is B anyways because fractional increase was asked.
Fractional increase = change in volume/old volume
The answer is b because that have seen the question in my pass question
