If the pressure of a gas contained in a close | vedclass

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

Question

(A) For a gas in a closed vessel,the volume $V$ remains constant. According to Gay-Lussac's Law,$\frac{P}{T} = 	ext{constant}$.Let the initial pressu

Vedclass · Accepted Answer

$250\, K$

Vedclass · Answer

(A) For a gas in a closed vessel,the volume $V$ remains constant. According to Gay-Lussac's Law,$\frac{P}{T} = 	ext{constant}$.Let the initial pressure be $P_1$ and the initial temperature be $T_1$ (in Kelvin).When the temperature is increased by $1\,^{\circ}C$,the new temperature is $T_2 = T_1 + 1$.The pressure increases by $0.4\%$,so the new pressure is $P_2 = P_1 + 0.004 P_1 = 1.004 P_1$.Using the relation $\frac{P_1}{T_1} = \frac{P_2}{T_2}$,we get:$\frac{P_1}{T_1} = \frac{1.004 P_1}{T_1 + 1}$$T_1 + 1 = 1.004 T_1$$1 = 1.004 T_1 - T_1$$1 = 0.004 T_1$$T_1 = \frac{1}{0.004} = 250\, K$.

If the pressure of a gas contained in a closed vessel is increased by $0.4\%$ when heated by $1\,^{\circ}C$,the initial temperature must be

Similar Questions

When the temperature of a gas in a closed vessel is increased by $2.4^{\circ} C$, its pressure increases by $0.5 \%$. The initial temperature of the gas is (in $^{\circ} C$)

Air pressure in a car tyre increases during driving. Explain.

For an ideal gas,the density of the gas is $\varrho_0$ when 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

At $0^{\circ}C$,the density of a fixed mass of a gas divided by pressure is $x$. At $100^{\circ}C$,the ratio will be

An open-mouthed container is filled with air at $60^{\circ}C$. The container is heated to a temperature $T$ such that $1/4$ of the air escapes. The value of $T$ is ........ $^{\circ}C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module