The expansion of unit mass of a perfect gas a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to Charles's Law,for a fixed mass of a gas at constant pressure,the volume $V$ is directly proportional to the absolute temperature $T$

Vedclass · Accepted Answer

$a =$ volume,$b = ^{\circ}C$ temperature

Vedclass · Answer

(A) According to Charles's Law,for a fixed mass of a gas at constant pressure,the volume $V$ is directly proportional to the absolute temperature $T$ (in Kelvin),i.e.,$V = kT$.However,when expressed in Celsius scale,$V = V_0(1 + \alpha t)$,where $t$ is the temperature in $^{\circ}C$.This equation represents a straight line $y = mx + c$,where $y$ is volume $(V)$ and $x$ is temperature ($t$ in $^{\circ}C$).In the given graph,the line intersects the horizontal axis at a negative value,which corresponds to $-273.15^{\circ}C$ (absolute zero).Since volume cannot be negative,the vertical axis $(a)$ must represent Volume $(V)$ and the horizontal axis $(b)$ must represent temperature in $^{\circ}C$ because the temperature scale in $^{\circ}C$ can have negative values.Therefore,$a =$ volume and $b = ^{\circ}C$ temperature.

The expansion of unit mass of a perfect gas at constant pressure is shown in the diagram. Here

Similar Questions

When air is pumped into a cycle tyre, the volume and pressure of the air in the tyre both increase. What about Boyle's law in this case?

The equation of state for $2 \ g$ of oxygen at a pressure $P$ and temperature $T$,when occupying a volume $V$,will be:

If a given mass of gas occupies a volume of $10\, cc$ at $1\, atm$ pressure and a temperature of $100^{\circ}C$ $(373.15\, K)$,what will be its volume at $4\, atm$ pressure,assuming the temperature remains constant?

At a given temperature,the pressure of an ideal gas of density $\rho$ is proportional to

At a temperature $T \ K$,the pressure of $4.0 \ g$ argon in a bulb is $p$. The bulb is put in a bath having a temperature higher by $50 \ K$ than the first one. $0.8 \ g$ of argon gas had to be removed to maintain the original pressure. The temperature $T$ is equal to ..... $K$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module