Question

A gas cylinder filled with argon is kept at a pressure of 177 atm and 25 degrees C. What is the gas pressure when the temperature of the cylinder and its contents are heated to 195 degrees C by exposure to fire?

Answer 1

Answer:

278 atm

Explanation:

We're gonna use this formula: $\frac{P_1}{T_1} =\frac{P_2}{T_2}$

P₁ = 177 atm

T₁ = 298 (Convert Celsius to kelvins by adding 273)

P₂ = ?

T₂ = 468 (195 + 273)

We got

$\frac{177}{298} =\frac{P_2}{468}$

We can cross-multiply or multiply both sides by 468

I'm gonna go with the latter

$(468)\frac{177}{298} =\frac{P_2}{468}(468)$

$P_2 = \frac{(468)(177)}{298}$

$P_2 = 277.973$

Answer 2

Answer:

$\boxed {\boxed {\sf 278 \ atm}}$

Explanation:

We are asked to find the gas pressure when the temperature of a gas is changed. We will use Gay-Lussac's Law, which states the pressure of a gas is proportional to the temperature of the gas. The formula for this law is:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Initially, the pressure is 177 atmospheres and the temperature is 25 degrees Celsius or 298 Kelvin.

$\frac {177 \ atm}{298 \ K}=\frac{P_2}{T_2}$

Then, the gas cylinder is exposed to fire and the temperature is raised to 195 degrees Celsius or 468 Kelvin, but the pressure is unknown.

$\frac {177 \ atm}{298 \ K }=\frac{P_2}{468 \ K}$

We are solving for the new pressure, so we must isolate the variable P₂. It is being divided by 468 Kelvin. The inverse operation of division is multiplication, so we multiply both sides of the equation by 468 Kelvin.

$468 \ K *\frac {177 \ atm}{298 \ K}=\frac{P_2}{468 \ K}*468 \ K$

$468 \ K *\frac {177 \ atm}{298 \ K}=P_2$

The units of Kelvin cancel.

$468 \ K *\frac {177 \ atm}{298 \ K }=P_2$

$468 * 0.593959731544 \ atm = P_2$

$277.973154362 \ atm = P_2$

The pressure in the cylinder after exposure to fire is approximately 278 atmospheres.