The temperature of a gas at -173^{circ} C is | vedclass

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

Question

(C) According to Gay-Lussac's Law,for a fixed mass of gas at constant volume,the pressure is directly proportional to its absolute temperature: $P \pr

Vedclass · Accepted Answer

$-73$

Vedclass · Answer

(C) According to Gay-Lussac's Law,for a fixed mass of gas at constant volume,the pressure is directly proportional to its absolute temperature: $P \propto T$.Therefore,$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$.Given:$P_{1} = 1 	ext{ atm}$$T_{1} = -173^{\circ} C = (-173 + 273) K = 100 K$$P_{2} = 2 	ext{ atm}$Substituting the values:$T_{2} = \frac{P_{2} 	imes T_{1}}{P_{1}} = \frac{2 	imes 100}{1} = 200 K$.Converting back to Celsius:$t_{2} = T_{2} - 273 = 200 - 273 = -73^{\circ} C$.

The temperature of a gas at $-173^{\circ} C$ is $1 \text{ atm}$ at constant volume. Up to what temperature should the gas be heated so that the pressure becomes $2 \text{ atm}$ (in $^{\circ} C$)?

Similar Questions

$A$ gas has a volume $V$ at a temperature of $27^{\circ}C$. If the temperature is increased to $327^{\circ}C$ while keeping the pressure constant,what will be the new volume of the gas?

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?

$A$ cylinder contains $10 \, kg$ of gas at a pressure of $10^7 \, N/m^2$. What amount of gas in $kg$ must be removed so that the final pressure becomes $2.5 \times 10^6 \, N/m^2$?

Explain why the pressure inside a tyre increases when a vehicle moves for a long period of time.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module