The mathematical form of Charles's law is given by $V_{t} = V_{0} \left( \frac{273.15 + t}{273.15} \right)$. Using this equation,explain the relationship between volume $(V)$ and temperature $(T)$ in Kelvin.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The given equation is $V_{t} = V_{0} \left( \frac{273.15 + t}{273.15} \right)$.
Let $T = 273.15 + t$,where $T$ is the temperature in Kelvin and $t$ is the temperature in degrees Celsius.
Substituting this into the equation,we get $V_{t} = V_{0} \left( \frac{T}{273.15} \right)$.
This can be rewritten as $V_{t} = \left( \frac{V_{0}}{273.15} \right) T$.
Since $V_{0}$ and $273.15$ are constants,we can write $V_{t} = k T$,where $k = \frac{V_{0}}{273.15}$.
This shows that volume $(V)$ is directly proportional to temperature $(T)$ in Kelvin,which is the definition of Charles's law.
The graph of $V$ versus $T$ (in Kelvin) is a straight line passing through the origin with a slope of $\frac{V_{0}}{273.15}$.

Explore More

Similar Questions

At constant pressure,the temperature of $20 \ L$ of helium gas is increased from $100 \ K$ to $300 \ K$. What will be the change in volume in $L$?

At $273 \, K$ temperature and $1 \, bar$ pressure,if the volume of a gas increases by $20 \%$,what temperature is required assuming pressure remains constant (in $, K$)?

When the temperature of a flask with a capacity of $1 \ L$ is increased from $25 \ ^oC$ to $35 \ ^oC$,calculate the volume of air in $mL$ that escapes from the flask.

The volume-temperature graphs of a given mass of an ideal gas at different constant pressures are shown below. What is the correct order of pressures?

In order to increase the volume of a gas by $10 \%$ at constant temperature,the pressure of the gas should be:

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