A gas thermometer is used as a standard thermometer for measurement of temperature. When the gas con

Q: Why must we use Kelvin in the gas law equation?

The direct proportionality between pressure and temperature ($P \propto T$) is a fundamental property of ideal gases that only holds true when temperature is measured on an absolute scale, like Kelvin. Zero Kelvin represents zero pressure (theoretically), whereas zero Celsius does not.

Q: Why is the triple point of water used as a reference?

The triple point of water is a unique and highly stable thermodynamic state where ice, liquid water, and water vapor coexist in equilibrium. It occurs at exactly 273.16 K and is used to define the Kelvin scale precisely.

Q: Can I just use Celsius and write $P_1 / t_1 = P_2 / t_2$?

Absolutely not! Pressure is not directly proportional to Celsius temperature. The correct linear relationship in Celsius is $P = P_0(1 + \alpha t)$, which doesn't allow for a simple ratio unless you convert to Kelvin first.

Q: Why did we subtract 273 instead of 273.15 to get Celsius?

While $0^\circ\text{C}$ is exactly 273.15 K, many classic textbook problems and older exam solutions use 273 as a convenient approximation. In this specific problem, subtracting 273 leads exactly to the intended answer of 45.68.

Q: What if the gas container expands when heated?

In a real, highly precise gas thermometer, the thermal expansion of the bulb must be accounted for. The volume wouldn't be strictly constant, and a correction factor involving the coefficient of volume expansion ($\gamma$) of the container material would be applied.

Question

A gas thermometer is used as a standard thermometer for measurement of temperature. When the gas container of the thermometer is immersed in water at its triple point $273.16	ext{ K}$, the pressure in the gas thermometer reads $3.0 	imes 10^4	ext{ N/m}^2$. When the gas container of the same thermometer is immersed in another system, the gas pressure reads $3.5 	imes 10^4	ext{ N/m}^2$. The temperature of this system is therefore ...... $^\circ	ext{C}$.

Sigma Percentile · Accepted Answer

### The Principle of Constant Volume

Imagine a rigid container filled with a fixed amount of gas. When we heat this container, the gas molecules gain kinetic energy and collide with the walls more forcefully. Because the container's volume cannot change, this increased force manifests entirely as an increase in pressure.

This fundamental behavior is the basis of a **constant volume gas thermometer**. According to Gay-Lussac's Law, for a fixed mass of an ideal gas kept at a constant volume, the pressure is directly proportional to its absolute temperature (measured in Kelvin). Mathematically, we express this as:

$$p \propto T$$

### Applying Gay-Lussac's Law

Because of this direct proportionality, the ratio of pressure to temperature remains constant. If we know the pressure and temperature at one state, we can easily find the temperature at another state if we know its pressure. We can set up the following ratio:

$$\frac{p_1}{T_1} = \frac{p_2}{T_2}$$

Rearranging this to solve for our unknown final temperature, $T_2$, we get:

$$T_2 = T_1 \left(\frac{p_2}{p_1}ight)$$

In our specific problem, the thermometer is first calibrated at the **triple point of water**, a highly stable reference point where water exists simultaneously as ice, liquid, and vapor. This occurs at exactly $T_1 = 273.16 	ext{ K}$. The pressure at this state is $p_1 = 3.0 	imes 10^4 	ext{ N/m}^2$. When placed in the new system, the pressure rises to $p_2 = 3.5 	imes 10^4 	ext{ N/m}^2$.

Substituting these values into our equation:

$$T_2 = 273.16 	imes \left( \frac{3.5 	imes 10^4}{3.0 	imes 10^4} ight)$$

The $10^4$ terms neatly cancel out, leaving us with a straightforward calculation:

$$T_2 = 273.16 	imes \left( \frac{3.5}{3.0} ight) = 318.68 	ext{ K}$$

### The Final Conversion Catch

We have successfully found the temperature of the new system, but it is in Kelvin. The question explicitly asks for the temperature in degrees Celsius.

To convert from Kelvin to Celsius, the standard formula is $T_{^\circ	ext{C}} = T_{	ext{K}} - 273.15$. However, in many classic physics problems (especially older JEE questions), it is a common convention to use the approximation of subtracting exactly $273$ unless high precision is demanded by the options. Following the intended logic of the problem's solution:

$$T_{^\circ	ext{C}} = 318.68 - 273$$

$$T_{^\circ	ext{C}} = 45.68^\circ	ext{C}$$

And there we have it! The temperature of the system is **$45.68^\circ	ext{C}$**.

Enter Numerical Value:

Visualized Solution

$V = constant$

$\frac{p _{1}}{T _{1}} = \frac{p _{2}}{T _{2}}$

$Substitution$

$T_{2} = 318.68 K$

$T_{^{\circ} C} = T_{K} - 273$

$What if V \neq = constant ?$

The Sigma Insight: Kinetic Theory of Gases

The Principle of Constant Volume

Applying Gay-Lussac's Law

The Final Conversion Catch

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​p1​​=T2​p2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​=318.68 K

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T∘C​=TK​−273

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if V=constant?