Question

1. What is the final volume of a 400.0 mL gas sample that is subjected to a
temperature change from 22.0 °C to 30.0 °C and a pressure change from 760.0 mmHg
to 360.0 mmHg?

Answer 1

Answer:

U will equate and make V2 the subject

Answer 2

867.3 mL is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0°C to 30.0°C and a pressure change from 760.0 mmHg to 360.0 mmHg.

What is Combined Gas Law ?

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's Law

It is expressed as

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

where

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Convert temperature Celsius into Kelvin

T(K) = T(°C) + 273

T₁(K) = 22 + 273

       = 295 K

T₂(K) = 30 + 273

        = 303 K

Now put the value in above formula we get

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

$\frac{760\ \text{mm Hg} \times 400\ mL}{295\ K} = \frac{360\ \text{mm Hg} \times V_2}{303\ K}$

$V_{2} = \frac{760\ \text{mm Hg} \times 400\ mL \times 303\ K}{295\ K \times 360\ \text{mm Hg}}$

V₂ = 867.3 mL

Thus from the above conclusion we can say that 867.3 mL is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0°C to 30.0°C and a pressure change from 760.0 mmHg to 360.0 mmHg.

Learn more about the Combined gas Law here: brainly.com/question/25587265

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