Question

1. A sample of gas in a balloon has an initial temperature of 40. ∘C and a volume of 1.88×10³ L . If the temperature changes to 84 ∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?

2.   What Celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 40. ∘C , V1= 1.88×10³ L ) to a volume of 3.76×10³ L ? Assume no change in pressure or the amount of gas in the balloon.

Answer 1

Answer – 1) We are given, T1 =40.0 ^oC + 273 = 313 K V1 = 1.88*10³ L ,

T2 = 84.0^oC + 273 = 357 K

We know Charles’s law-

V1/T1 = V2/T2

So, V2 = V1 * T2 / T1

            = 1.88*10³ L * 357 K / 313 K

            = 2.14*10³ L

The new volume of the gas is 2.14*10³ L

2) We are given, T1 =40.0 ^oC + 273 = 313 K V1 = 1.88*10³ L ,

T2 = ? , V2 = 3.76*10³ L

We know Charles’s law-

V1/T1 = V2/T2

T2 = T1 * V2 / V1

     = 313 K * 3.76*10³ L / 1.88*10³ L

      = 626 K

So, T2 = 626-273 = 353 K

The degree Celsius temperature, T2, is required to change the volume of the gas sample in Part A is 353 K