Question

< Week 11 Review Section 8.3-8.8 (60) Temperature and Volume Il Review Charles's law states that the volume of a gas is directly related to the absolute temperature when there is no change in the pressure or amount of gas: (Figure 1) Part A A sample of gas in a balloon has an initial temperature of 19 °C and a volume of 1.05x10° L. If the temperature changes to 73 °C, and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas? Express your answer with the appropriate units. View Available Hint(3) V2 = 6881.71 Submit Previous Answers X Incorrect; Try Again; 2 attempts remaining Part B Figure < 1 of 1 > What Celsius temperature. Ty, is required to change the volume of the gas sample in Part A Ti = 19 °C, Vi= 1.05x10 L ) to a volume of 2.10x10'L ? Assume no change in pressure or the amount of gas in the balloon. Express your answer with the appropriate units. View Available Hint(s) T2 = Value Units Submit Lower temperature Smaller volume Higher temperature Larger volume Provide Feedback

Answer 1

Part A:

According to Charles' law, the volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

V$\alpha$ T

=> V/T = constant

=> V₁ / T₁ = V₂ / T₂ ------------------------------------ (1)

Given the initial volume V₁ = 1.05 x 10³ L

Initial temperature T₁ = 19⁰C =(19+273) = 292K

Final temperature T₂ = 73⁰C = (73+ 273)K = 346K

Final volume V₂ = ?

Substitute this values in equation (1), we get

1.05 x 10³ L / 292K = V₂ / 346K

=> V₂ = [1.05 x 10³ L / 292K ] x 346K

=> V₂ = 1244.1 L = 1.24 x 10³ L

Hence, final volume = 1.24 x 10³ L

Part B :

According to Charles' law, the volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

V$\alpha$ T

=> V/T = constant

=> V₁ / T₁ = V₂ / T₂ ------------------------------------ (1)

Given the initial volume V₁ = 1.05 x 10³ L

Initial temperature T₁ = 19⁰C =(19+273) = 292K

Final temperature T₂ = ?

Final volume V₂ = 2.10 x 10³ L

Substitute this values in equation (1), we get

1.05 x 10³ L / 292K = 2.10 x 10³ L/ T₂

=> T₂ = [2.10 x 10³ L x 292K ] / 1.05 x 10³ L

=> T₂ = 584 K

Hence, final temperature = (584 -273) ⁰ C

= 311⁰C