Question

A sample of gas in a balloon has an initial temperature of 11 ∘C and a volume of 1.03×103 L . If the temperature changes to 70. ∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?

Answer 1

Answer:

The answer to your question is  V2 = 1244 L

Explanation:

Data

Temperature 1 = 11°C

Volume = 1.03 x 10³ L

Temperature 2 = 70°C

Volume 2 = ?

To solve this problem use the Charles' law

        V1/ T1 = V2 / T2

-Solve for V2

        V2 = V1T2 / T1

Process

1.- Convert temperature to °K

Temperature 1 = 11°C + 273 = 284°K

Temperature 2 = 70°C + 273 = 343°K

2.- Substitution

         V2 = (1.03 x 10³)(343) / 284

3.- Simplification

        V2 = 353290 / 284

4.- Result

        V2 = 1244 L

Answer 2

Answer:

The new volume of the gas is 1244 L

Explanation:

Step 1: Data given

The initial temperature in the balloon = 11.0 °C = 284 K

The initial volume in the balloon is 1.03 *10³ L = 1030 L

The temperature increases to 70.0 °C = 343 K

The pressure stays constant

Step 2: Calculate the new volume of the gas

V1/T1 = V2/T2

⇒with V1 = the initial volume of the gas = 1030 L

⇒with T1 = the initial temperature in the balloon = 284 K

⇒with V2 = the new volume of the gas = TO BE DETERMINED

⇒with T2 = the new temperature in the balloon = 343 K

1030 L / 284 K = V2 / 343 K

V2 = 1244 L

The new volume of the gas is 1244 L