
The pressure of a gas filled in the bulb of a constant volume gas thermometer at $0^{\circ}C$ and $100^{\circ}C$ are \[28.6cm\]and \[36.6cm\] of mercury respectively. The temperature of bulb at which pressure will be \[35cm\]of mercury will be:
\[\begin{align}
& A{{.80}^{\circ }}C \\
& B{{.70}^{\circ }}C \\
& C{{.55}^{\circ }}C \\
& D{{.40}^{\circ }}C \\
\end{align}\]
Answer
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Hint: A thermometer is an instrument which is used to measure the heat of any given body. It works on the principle of the zeroth law of thermodynamics. The zeroth law of thermodynamics states that, if two systems $A$ and $B$ are in thermal equilibrium with a third system $C$ then, the system $A$ and $B$ are said to be in thermal equilibrium with each other.
Formula used:
$\dfrac{T-T_{1}}{T_{2}-T_{1}}=\dfrac{P-P_{1}}{P_{2}-P_{1}}$
Complete answer:
We know that a constant volume gas thermometer is a thermometer whose bulb contains some fixed amount of dilute mercury. Mercury is used in the thermometer because of its high coefficient of expansion and it is one of the liquids which is in liquid state at room temperature.
From Gay-Lussac’s law of gases, we know that at constant volume, the pressure of an ideal gas is directly proportional to the temperature at which the pressure is measured.
Let us assume that the mercury gas in the thermometer is ideal in nature. Then from Gay-Lussac’s law of gases, $P\propto T$, then we can also say that $\Delta P\propto \Delta T$
Here it is given that, $P_{1}=28.6cm$ at $T_{1}=0^{\circ}C$ and $P_{2}=36.6$ at $T_{2}=100^{\circ}C$.
Let $P=35cm$ at $T$
Let us take the $T_{1}=0^{\circ}$ as the reference, then we can say that $T-T_{1}\propto P-P_{1}$ similarly, $T_{2}-T_{1}\propto P_{2}-P_{1}$
Taking the ratio we get $\dfrac{T-T_{1}}{T_{2}-T_{1}}=\dfrac{P-P_{1}}{P_{2}-P_{1}}$
Substituting we get, $\dfrac{T-0}{100-0}=\dfrac{35-28.6}{36.6-28.6}$
On solving we get, $T=\dfrac{35-28.6}{36.6-28.6}\times 100^{\circ}C=\dfrac{6.4}{8}\times 100^{\circ}C=0.8\times 100^{\circ}C=80^{\circ}C$
So, the correct answer is “Option A”.
Note:The constant volume mercury thermometer is the normal thermometer which is used by doctors to measure the body temperature of a patient. However, the same is used for experiments also, but the scale of the thermometer may vary depending on the need of the experiment
Formula used:
$\dfrac{T-T_{1}}{T_{2}-T_{1}}=\dfrac{P-P_{1}}{P_{2}-P_{1}}$
Complete answer:
We know that a constant volume gas thermometer is a thermometer whose bulb contains some fixed amount of dilute mercury. Mercury is used in the thermometer because of its high coefficient of expansion and it is one of the liquids which is in liquid state at room temperature.
From Gay-Lussac’s law of gases, we know that at constant volume, the pressure of an ideal gas is directly proportional to the temperature at which the pressure is measured.
Let us assume that the mercury gas in the thermometer is ideal in nature. Then from Gay-Lussac’s law of gases, $P\propto T$, then we can also say that $\Delta P\propto \Delta T$
Here it is given that, $P_{1}=28.6cm$ at $T_{1}=0^{\circ}C$ and $P_{2}=36.6$ at $T_{2}=100^{\circ}C$.
Let $P=35cm$ at $T$
Let us take the $T_{1}=0^{\circ}$ as the reference, then we can say that $T-T_{1}\propto P-P_{1}$ similarly, $T_{2}-T_{1}\propto P_{2}-P_{1}$
Taking the ratio we get $\dfrac{T-T_{1}}{T_{2}-T_{1}}=\dfrac{P-P_{1}}{P_{2}-P_{1}}$
Substituting we get, $\dfrac{T-0}{100-0}=\dfrac{35-28.6}{36.6-28.6}$
On solving we get, $T=\dfrac{35-28.6}{36.6-28.6}\times 100^{\circ}C=\dfrac{6.4}{8}\times 100^{\circ}C=0.8\times 100^{\circ}C=80^{\circ}C$
So, the correct answer is “Option A”.
Note:The constant volume mercury thermometer is the normal thermometer which is used by doctors to measure the body temperature of a patient. However, the same is used for experiments also, but the scale of the thermometer may vary depending on the need of the experiment
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