Question

6. A gas thermometer is constructed of two gas-containing bulbs, each in a water bath. The pressure difference
between the two bulbs is measured by a mercury manometer as shown. Appropriate reservoirs, not shown in the diagram,
maintain constant gas volume in the two bulbs. There is no difference in pressure when both baths are at the triple point of
water. The pressure difference is 120 torr, when one bath is at the triple point and the other is at the boiling point of water.
It is 90.0 torr when one bath is at the triple point and the other is at an unknown temperature to be measured. What is the
unknown temperature?​

Answer 1

Given :

A gas thermometer is constructed of two gas-containing bulbs, each in a water bath.

The pressure difference is 120 torr, when one bath is at the triple point and the other is at the boiling point of water.

So for

Step1:

In this step

Let us consider that in the left hand thermometer, $T_L$ be the temperature and $P_L$ be the pressure.

And now similarly for right hand thermometer $T_R$ be the temperature and $P_R$ be the pressure.

So now according to our problem

The pressure is the same in the two thermometers when they are both at the triple point of water. We take this pressure to be$P_3$

So for each thermometer the writing equation is

$T_L=(273.16 K)(P_1/P_3) and T_R=(273.16 K)(P_g/P_3)$

Now after subtracting we get

$T_L-T_R=(273.16 K)(P_L-P_g/P_3)$

Step2:

In this step2

We will take $T_R=273.16K, T_L=373.125K\\Then \\P_L-P_R=120torr$

So after solving we get

$373.125K-273.16K=(273.16K)(120 torr/P_3)$

So the result is

$P_3=328 torr.$

As we know that

$T_R$ be the unknow temperature.

So the pressure difference is given by

$P_L-P_R=90.0 torr$

Step 3 :

In this step

We will solve the the pressure difference as

$273.16K-T_R=(273.16K)(90.0 torr/328 torr)$

So we get

$T_R=348K.$

Hence, we obtain the unknown temperature as$T_3=348K.$