Question

the pressure of gas in constant volume thermometer are 80cm,109.3cm,94.65cm of Hg at theice point,steam point and in a hot bath respectively.
calculate temperature of the bath
​

Answer 1

Answer:

Since the volume of gas in thermometer is constant,

T

1

P

1

=

T

2

P

2

Since temperature of melting ice is 273.15K,

273.15

80

=

T

2

160

T

2

=2(273.15)K=546.30K

Answer is 546.30K

Answer 2

Given :-

Pressure of the gas at ice point = 80 cm

Pressure of the gas at steam point = 90 cm

Pressure of the gas in a heated wax bath = 100 cm

To Find :-

Temperature of the wax bath.

Solution :-

We know that,

• p = Pressure
• t = Temperature

According to the question,

$\sf P_{ice} =$ Pressure at ice point = 80 cm of Hg

$\sf T_{ice} =$ Ice point temperature = 0°C

$\sf P_{steam}=$ Pressure at steam point = 90 cm of Hg

$\sf T_{steam} =$ Steam point temperature = 100°C

Pressure of wax bath = P = 100cm of Hg

Let “T” be the temperature of the wax bath.

Now,

$\sf \dfrac{(P-P_{ice})}{(P_{steam}-P_{ice})} =\dfrac{(T-T_{ice})}{(T_{steam}-T_{ice})}$

Substituting their values, we get

$\sf T=\dfrac{100 \ cm-80 \ cm}{90 \ cm-80 \ cm} \times 100^{o} \: C$

$\sf T=200^{o} \: C$

Therefore, the temperature of the wax bath is 200° C