The pressure (P) and temperature (T) relation | vedclass

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

Question

(D) Given the relation $PT^2 = C$ (where $C$ is a constant).Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.Substituting $P$ into

Vedclass · Accepted Answer

$\frac{3}{T}$

Vedclass · Answer

(D) Given the relation $PT^2 = C$ (where $C$ is a constant).Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.Substituting $P$ into the given equation: $\left(\frac{nRT}{V}ight)T^2 = C$.This simplifies to $\frac{nRT^3}{V} = C$, which implies $V = \left(\frac{nR}{C}ight)T^3$.Let $K = \frac{nR}{C}$, so $V = KT^3$.The volume expansion coefficient $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.Differentiating $V$ with respect to $T$: $\frac{dV}{dT} = 3KT^2$.Now, substitute into the formula for $\gamma$: $\gamma = \frac{1}{KT^3} 	imes (3KT^2) = \frac{3}{T}$.

The pressure $(P)$ and temperature $(T)$ relationship of an ideal gas obeys the equation $PT^2 = \text{constant}$. The volume expansion coefficient of the gas will be:

Similar Questions

Nitrogen gas $N_2$ of mass $28 \,g$ is kept in a vessel at a pressure of $10 \,atm$ and temperature $57^{\circ} C$. Due to leakage of $N_2$ gas,its pressure falls to $5 \,atm$ and temperature to $27^{\circ} C$. The amount of $N_2$ gas leaked out is .......... $g$.

On $0^{\circ}C$,the pressure measured by a barometer is $760 \, mm$. What will be the pressure at $100^{\circ}C$?

Absolute zero $(0\, K)$ is that temperature at which

For an ideal gas,the density of the gas is $\rho_0$ when the temperature and pressure of the gas are $T_0$ and $P_0$ respectively. When the temperature of the gas is $2 T_0$,its pressure becomes $3 P_0$. The new density will be:

The product of the pressure and volume of an ideal gas is

Vedclass Test Series

Exam Paper Generator

Online Exam Module