
The pressure of a definite mass of a gas is directly proportional to the temperature and inversely proportional to the volume under the given conditions. If temperature is increased by 40% and volume is decreased by 20% then the new pressure will be?
(a) Be increased by 75%
(b) Reduce to 25%
(c) Be increased by 20%
(d) Increased by 28%
Answer
626.1k+ views
Hint: Here, we will use the definition of percentage to find the new value of the pressure of the given mass of gas. Then, we will find the difference between the new value and the original value to find the percent change.
Complete step-by-step answer:
Since, we have been given that the pressure of the definite amount of gas is directly proportional to the temperature and inversely proportional to the volume.
Let us denote the pressure, temperature and volume by P, T and V respectively.
So, we can write:
\[P\propto \,\dfrac{T}{V}\]
\[P=K\dfrac{T}{V}.............\left( 1 \right)\] (K is a proportionality constant)
Since, the temperature has increased by 40%. So, the new temperature is:
$\begin{align}
& T'=T+\dfrac{40}{100}T \\
& T'=T+\dfrac{2}{5}T \\
& T'=\dfrac{7T}{5} \\
\end{align}$
Also, the volume is decreased by 20%. So, the new volume is:
$\begin{align}
& V'=V-\dfrac{20}{100}V \\
& V'=V-\dfrac{V}{5} \\
& V'=\dfrac{4V}{5} \\
\end{align}$
On substituting the new values of T and V in equation (1), we get the new value of pressure as:
$\begin{align}
& P'=K\dfrac{T'}{V'} \\
& \Rightarrow P'=K\dfrac{\dfrac{7T}{5}}{\dfrac{4V}{5}} \\
& \Rightarrow P'=K\dfrac{7T}{4V} \\
& \Rightarrow P'=\dfrac{7}{4}.\dfrac{KT}{V} \\
& \Rightarrow P'=\dfrac{7}{4}P \\
\end{align}$
So, the new value of the pressure of definite mass of the given gas is seven-fourth of the original value of the pressure.
Percentage change in the pressure of the gas is:
$\begin{align}
& =\dfrac{P'-P}{P}\times 100 \\
& =\dfrac{\left( \dfrac{7P}{4}-P \right)}{P}\times 100 \\
& =\dfrac{3P}{4P}\times 100 \\
& =75 \\
\end{align}$
So, the value of percent change in the pressure of the gas is 75%.
Hence, option (a) is the correct answer.
Note: Students should note that the final value of pressure is seven-fourth of the initial value of the pressure. It means that the final value of pressure is greater than the initial value and hence, it denotes an increase in the percentage value.
Complete step-by-step answer:
Since, we have been given that the pressure of the definite amount of gas is directly proportional to the temperature and inversely proportional to the volume.
Let us denote the pressure, temperature and volume by P, T and V respectively.
So, we can write:
\[P\propto \,\dfrac{T}{V}\]
\[P=K\dfrac{T}{V}.............\left( 1 \right)\] (K is a proportionality constant)
Since, the temperature has increased by 40%. So, the new temperature is:
$\begin{align}
& T'=T+\dfrac{40}{100}T \\
& T'=T+\dfrac{2}{5}T \\
& T'=\dfrac{7T}{5} \\
\end{align}$
Also, the volume is decreased by 20%. So, the new volume is:
$\begin{align}
& V'=V-\dfrac{20}{100}V \\
& V'=V-\dfrac{V}{5} \\
& V'=\dfrac{4V}{5} \\
\end{align}$
On substituting the new values of T and V in equation (1), we get the new value of pressure as:
$\begin{align}
& P'=K\dfrac{T'}{V'} \\
& \Rightarrow P'=K\dfrac{\dfrac{7T}{5}}{\dfrac{4V}{5}} \\
& \Rightarrow P'=K\dfrac{7T}{4V} \\
& \Rightarrow P'=\dfrac{7}{4}.\dfrac{KT}{V} \\
& \Rightarrow P'=\dfrac{7}{4}P \\
\end{align}$
So, the new value of the pressure of definite mass of the given gas is seven-fourth of the original value of the pressure.
Percentage change in the pressure of the gas is:
$\begin{align}
& =\dfrac{P'-P}{P}\times 100 \\
& =\dfrac{\left( \dfrac{7P}{4}-P \right)}{P}\times 100 \\
& =\dfrac{3P}{4P}\times 100 \\
& =75 \\
\end{align}$
So, the value of percent change in the pressure of the gas is 75%.
Hence, option (a) is the correct answer.
Note: Students should note that the final value of pressure is seven-fourth of the initial value of the pressure. It means that the final value of pressure is greater than the initial value and hence, it denotes an increase in the percentage value.
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