Question

1. A sample of gas in a balloon has an initial temperature of
40. ∘C and a volume of 1.88×10^{3} L . If the temperature
changes to 84 ∘C , and there is no change of pressure or amount of
gas, what is the new volume, *V*2, of the gas?

2. What Celsius temperature, *T*2, is
required to change the volume of the gas sample in Part A
(*T*1 = 40. ∘C , *V*1= 1.88×10^{3} L ) to a
volume of 3.76×10^{3} L ? Assume no change in pressure or
the amount of gas in the balloon.

Answer #1

**Answer – 1)** We are given, T1 =40.0
^{o}C + 273 = 313 K V1 = 1.88*10^{3} L ,

T2 = 84.0^{o}C + 273 = 357 K

We know Charles’s law-

V1/T1 = V2/T2

So, V2 = V1 * T2 / T1

= 1.88*10^{3} L * 357 K / 313 K

= 2.14*10^{3} L

The new volume of the gas is 2.14*10^{3} L

**2)** We are given, T1 =40.0 ^{o}C + 273 =
313 K V1 = 1.88*10^{3} L ,

T2 = ? , V2 = 3.76*10^{3} L

We know Charles’s law-

V1/T1 = V2/T2

**T2 = T1 * V2 / V1**

** =** 313 K *
3.76*10^{3} L / 1.88*10^{3} L

= 626 K

So, T2 = 626-273 = 353 K

The degree Celsius temperature, *T*2, is required to
change the volume of the gas sample in Part A is **353
K**

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