Question

7. A sample of argon gas has a volume of 0.43 mL at 24 C. At what temperature in Celsius
will it have a volume of 0.57 mL?

Answer 1

Answer:

120.55 °C

Explanation:

Given data:

Initial volume of gas = 0.43 mL

Final volume of gas = 0.57 mL

Initial temperature = 24°C (24+273 = 297 k)

Final temperature = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁  

T₂ = 0.57 mL × 297 K / 0.43 mL

T₂ = 169.29 mL.K / 0.43 mL

T₂ = 393.7 K

Temperature in celsius:

393.7 K - 273.15 = 120.55 °C

Answer 2

The new temperature will be "120.55°C". To understand the calculation, check below.

Volume and Temperature

According to the question,

Initial volume, V₁ = 0.43 mL

Final volume, V₂ = 0.57 mL

Initial Temperature, T₁ = 24°C or,

                                     = 297 K

We know the relation,

→ $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Or,

The final temperature be:

   T₂ = $\frac{V_2\times T_1}{V_1}$

By substituting the values, we get

        = $\frac{0.57\times 297}{0.43}$

        = $\frac{169.29}{0.43}$

        = 393.7 K or,

        = 120.55°C

Thus the above approach is correct.

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