Question

​29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.
At what temperature would the gas occupy 1.30 L at
210.0 kPa?

Answer 1

The answer for the following mention bellow.

• Therefore the final temperature of the gas is 260 k

Explanation:

Given:

Initial pressure ($P_{1}$) = 150.0 kPa

Final pressure ($P_{2}$) = 210.0 kPa

Initial volume ($V_{1}$) = 1.75 L

Final volume ($V_{2}$) = 1.30 L

Initial temperature ($T_{1}$) = -23°C = 250 k

To find:

Final temperature ($T_{2}$)

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas  constant

T represents the temperature of the gas

We know;

$\frac{P*V}{T}$ = constant

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

Where;

($P_{1}$) represents the initial pressure of the gas

($P_{2}$) represents the final pressure of the gas

($V_{1}$) represents the initial volume of the gas

($V_{2}$) represents the final volume of the gas

($T_{1}$) represents the initial temperature of the gas

($T_{2}$) represents the final temperature of the gas

So;

$\frac{150 * 1.75}{210 * 1.30}$ = $\frac{260}{T_{2} }$

($T_{2}$) =260 k

Therefore the final temperature of the gas is 260 k