11) A fixed amount of gas at 25.0 °C occupies a volume of 8.66 L when the pressure is 629 torr. Use Charles's law to calculate the volume (L) the gas will occupy when the temperature is increased to
112 °C while maintaining the pressure at 629 torr.
A) 1.93
B) 38.8
C) 11.2
D) 9.26
E) 6.70

Option C. 11.2 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 25.0 °C

Initial volume (V₁) = 8.66 L

Final temperature (T₂) = 112 °C

Pressure = constant

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 25 °C

Initial temperature (T₁) = 25 °C +273

Initial temperature (T₁) = 298 K

Final temperature (T₂) = 112 °C

Final temperature (T₂) = 112 °C + 273

Final temperature (T₂) = 385 K

Finally, we shall determine final volume of the gas. This can be obtained as follow:

Initial temperature (T₁) = 298 K

Initial volume (V₁) = 8.66 L

Final temperature (T₂) = 385 K

Pressure = constant

Final volume (V₂) =?

V₁/T₁ = V₂/T₂

8.66/298 = V₂/385

Cross multiply

298 × V₂ = 8.66 × 385

298 × V₂ = 3334.1

Divide both side by 298

V₂ = 3334.1 / 298

V₂ = 11.2 L

Thus, the final (i.e new) volume of the gas is 11.2 L

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

14,16 18 reacts with group