A gas cylinder can withstand a pressure of 15 | vedclass

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

Question

(B) According to Gay-Lussac's Law,for a fixed amount of gas at constant volume,$P \propto T$ or $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Given: $P_1 = 12 \

Vedclass · Accepted Answer

$102$

Vedclass · Answer

(B) According to Gay-Lussac's Law,for a fixed amount of gas at constant volume,$P \propto T$ or $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Given: $P_1 = 12 \ atm$,$T_1 = 27 \ ^oC = 27 + 273 = 300 \ K$.Maximum pressure the cylinder can withstand,$P_2 = 15 \ atm$.Substituting the values: $\frac{12}{300} = \frac{15}{T_2}$.$T_2 = \frac{15 	imes 300}{12} = 375 \ K$.Converting back to Celsius: $T_2( ^oC) = 375 - 273 = 102 \ ^oC$.Thus,the cylinder will not burst up to $102 \ ^oC$.

$A$ gas cylinder can withstand a pressure of $15 \ atm$. At $27 \ ^oC$,the pressure in the cylinder is $12 \ atm$. Up to what temperature in $^oC$ will the cylinder not burst?

Similar Questions

$A$ gas occupies $10 \ L$ at $0 \ ^{\circ}C$ and $2 \ atm$ pressure. At what temperature will the pressure in the container be $2.5 \ atm$,assuming the volume remains constant (in $^{\circ}C$)?

The minimum pressure required to compress $600 \; dm^{3}$ of a gas at $1 \; bar$ to $150 \; dm^{3}$ at $40^{\circ} C$ is ......... $bar$.

The correct value of the gas constant $R$ is close to

The following graphs are based on Boyle's Law at constant mass and temperature. Describe what each graph represents.

If $P$ is pressure and $\rho$ is density of a gas,then $P$ and $\rho$ are related as

Vedclass Test Series

Exam Paper Generator

Online Exam Module