3. Let T

L

be the temperature and p

L

be the pressure in the left-hand thermometer. Similarly, let T

R

be the temperature and p

R

be the pressure in the right-hand thermometer. According to the problem

statement, 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

. Writing Eq. 19-5 for each thermometer,

T

L

= (273.1 6 K)

p

L

p

3



and

T

R

= (273.1 6 K)

p

R

p

3



,

we subtract the second equation from the first to obtain

T

L

− T

R

= (273.1 6 K)

p

L

− p

R

p

3



.

First, we take T

L

= 373.125 K (the boiling point of water) and T

R

= 273.16 K (the triple point of water).

Then, p

L

− p

R

= 120 torr. We solve

373.125 K

− 273.16 K = (273.1 6 K)

120 torr

p

3



for p

3

. The result is p

3

= 328 torr. Now, we let T

L

= 273.16 K (the triple point of water) and T

R

be the

unknown temperature. The pressure difference is p

L

− p

R

= 90.0 torr. Solving

273.1 6 K

− T

R

= (273.1 6 K)

90.0 torr

328 torr



for the unknown temperature, we obtain T

R

= 348 K.