Question

450 ml of oxygen gas at 20°C is heated to 50°C. What is the new volume of the gas at constant pressure?

Answer 1

$\text{\underline{\underline{Initial\:conditions:}}}$

$\sf{V_{1}}$ = 450 mL

$\sf{T_{1}}$ = 20° C

$\sf{T_{1}}$ = 20 + 273

$\sf{T_{1}}$ = $\boxed{293 K}$


$\text{\underline{\underline{Final\:conditions:}}}$

$\sf{V_{2}}$ = ?

$\sf{T_{2}}$ = 50° C

$\sf{T_{2}}$ = 50 + 273

$\sf{T_{2}}$ = $\boxed{323 K}$


$\text{\underline{\underline{According\:to\:Charle's Law}}}$:

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


Now, substituting the values, in the above formula, we get:

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

$\sf{ \frac{450}{293} = \frac{V_{2}}{323}}$

$\sf{V_{2} = \frac{323 \times 450}{293}}$

$\sf{V_{2} = \frac{145350}{293}}$

$\sf{V_{2} = \boxed{496.0 \: mL}}$


$\text{\underline{\underline{Therefore:}}}$

The new volume of the gas at constant pressure is 496.0 mL.


$\text{\underline{\underline{Note\:on\:Charle's\:Law}}}$ :


$\bullet$ In 1787, French scientist, Jacques Charles studied the effect of change of temperature on the volume of a fixed amount of gas at constant pressure.


$\bullet$ He observed that for a given mass of a gas, the volume increases with increase in temperature and vice-versa at constant pressure.


$\bullet$ Then he generalised his observations in the form of a law called Charle's Law.


$\bullet$ Charle's Law states that the volume of a given mass of a gas increases or decreases by $\frac{1}{273.15}$ of the volume at 0°C for each degree rise or fall in temperature respectively, provided pressure is kept constant.

Answer 2

Answer:

Final volume will be 496 mL.

Explanation:

Given that -

• Initial volume (V1) = 450 mL
• Initial temperature (T1) = 20°C
• Final temperature (T2) = 50°C

To find -

• Final volume (V2)

Solution -

Let's change the temperature into Kelvin.

T1 = 20°C

→ T1 = 20 + 273

→ T1 = 293 K

T2 = 50°€C

→ T2 = 50 + 273

→ T2 = 323 K

Using Charle's law,

→ V1/T1 = V2/T2

→ 450/293 = V2/323

→ V2 = 496 mL

Hence, final volume will be 496 mL.