Two ideal gas thermometers $A$ and $B$ use oxygen and hydrogen respectively. The following observations are made:
Temperature Pressure thermometer $A$ Pressure thermometer $B$
Triple-point of water $1.250 \times 10^{5} \; Pa$ $0.200 \times 10^{5} \; Pa$
Normal melting point of sulphur $1.797 \times 10^{5} \; Pa$ $0.287 \times 10^{5} \; Pa$

$(a)$ What is the absolute temperature of the normal melting point of sulphur as read by thermometers $A$ and $B$?
$(b)$ What do you think is the reason behind the slight difference in answers of thermometers $A$ and $B$? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) For thermometer $A$:
At triple point of water,$T = 273.16 \; K$,$P_A = 1.250 \times 10^{5} \; Pa$.
At melting point of sulphur,$P_1 = 1.797 \times 10^{5} \; Pa$.
Using Charles' Law,$T_1 = (P_1 / P_A) \times 273.16 = (1.797 / 1.250) \times 273.16 = 392.69 \; K$.
For thermometer $B$:
At triple point of water,$T = 273.16 \; K$,$P_B = 0.200 \times 10^{5} \; Pa$.
At melting point of sulphur,$P_2 = 0.287 \times 10^{5} \; Pa$.
Using Charles' Law,$T_1 = (P_2 / P_B) \times 273.16 = (0.287 / 0.200) \times 273.16 = 391.98 \; K$.
$(b)$ The gases oxygen and hydrogen are not perfectly ideal. The discrepancy arises because real gases deviate from ideal gas behavior. To reduce the discrepancy,the experiment should be performed at lower pressures,where gases behave more like ideal gases.

Explore More

Similar Questions

The value of critical temperature in terms of van der Waals' constants $a$ and $b$ is given by

For a van der Waals gas,if $P_c, V_c$,and $T_c$ are the critical pressure,volume,and temperature respectively,then the value of $P_cV_c/T_c$ is:

Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at $STP$. Take the diameter of an oxygen molecule to be $3 \mathring A$.

There are two identical chambers,completely thermally insulated from surroundings. Both chambers have a partition wall dividing the chambers into two compartments. Compartment $1$ is filled with an ideal gas and Compartment $3$ is filled with a real gas. Compartments $2$ and $4$ are vacuum. $A$ small hole (orifice) is made in the partition walls and the gases are allowed to expand into the vacuum.
Statement $-1$: No change in the temperature of the gas takes place when an ideal gas expands in a vacuum. However,the temperature of a real gas goes down (cooling) when it expands in a vacuum.
Statement $-2$: The internal energy of an ideal gas is only kinetic. The internal energy of a real gas is kinetic as well as potential.

If the pressure of a real gas $O_{2}$ in a container is given by $P = \frac{RT}{2V - b} - \frac{a}{4b^{2}}$, then the mass of the gas in the container is: (in $\text{ g}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo