Question

A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the gas changes to 4 atm?

Answer 1

The answer for the following problem is mentioned below.

• Therefore the final temperature of the gas is 740 K

Explanation:

Given:

Initial pressure of the gas ($P_{1}$) = 1.8 atm

Final pressure of the gas ($P_{2}$)  = 4 atm

Initial temperature of the gas ($T_{1}$) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas ($T_{2}$)

We know;

From the ideal gas equation;

we know;

P  × V = n × R × T

So;

we can tell from the above equation;

    P ∝ T

(i.e.)

      $\frac{P}{T}$ = constant

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

Where;

$P_{1}$  = initial pressure of a gas

$P_{2}$ = final pressure of a gas

$T_{1}$ = initial temperature of a gas

$T_{2}$ = final temperature of a gas

        $\frac{1.8}{4}$ = $\frac{333}{T_{2} }$

   $T_{2}$ =$\frac{333*4}{1.8}$

    $T_{2}$ = 740 K

Therefore the final temperature of the gas is 740 K