Question

A gas is trapped in a 56 liter vessel. The temperature inside is 234 Celsius and thepressure is 1.1 atmospheres. What is the new pressure if the volume is compressed to54 liters and the temperature is lowered to 200 ​o​C?

Answer 1

Answer:

The new pressure is 1.06 atmospheres

Explanation:

We can calculate the new pressure from using the Combined gas law equation.

The combined gas law equation is

$\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }$

Where $P_{1}$ is the initial pressure

$V_{1}$ is the initial volume

$T_{1}$ is the initial temperature

$P_{2}$ is the final pressure

$V_{2}$ is the final volume

and $T_{2}$ is the final temperature

From the question,

$V_{1}$ = 56 L

$T_{1}$ = 234 °C = (234 + 273.15)K = 507.15K

$P_{1}$ = 1.1 atm

$V_{2}$ = 54 L

$T_{2}$ = 200 °C = (200 + 273.15)K = 473.15K

Now, to determine $P_{2}$ (the new pressure), we will put the given values into the equation

$\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }$

$\frac{1.1 \times 56}{507.15}= \frac{P_{2} \times 54 }{473.15}$

∴ $P_{2} = \frac{1.1 \times 56 \times 473.15 }{54 \times 507.15}$

Then,

$P_{2} = \frac{29146.04 }{27386.1}$

$P_{2} = 1.06$ atm

Hence, the new pressure is 1.06 atmospheres.