A sample of gas in a balloon has an initial temperature of 9.0∘C and a volume of 1.75×103L . If the temperature changes to 80.∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?

87

Not sure how you got 87! :)

Start with PV/T=constant, where T is in °K.
V1=1.75*10³L
T1=(9.0+273.15)=282.15°K
T2=(80+273.15)=353.15°K
(ok to use excess significant digits for calculation, but reduce answer to appropriate number).
since Pressure & amount of gas remain constant, we have
V/T=constant, or
V1/T1=V2/T2
=>
V2=V1(T2/T1)
=1.75*10³(353.15/282.15)
=2.19*10³ L

Not sure how you got 87 as well. However, 2.19*103L or 2190 L is correct!

To find the new volume (V2) of the gas, we can use Charles' Law, which states that the volume of a gas is directly proportional to its temperature when pressure and amount of gas remain constant.

The formula for Charles' Law is:

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.

Let's plug in the values given in the question:

V1 = 1.75×10^3 L
T1 = 9.0∘C = 9.0 + 273.15 K (converting Celsius to Kelvin)
T2 = 80.∘C = 80.0 + 273.15 K

Now we can substitute these values into the equation:

(1.75×10^3 L)/(9.0 + 273.15 K) = V2/(80.0 + 273.15 K)

To solve for V2, we can cross-multiply and then divide:

(1.75×10^3 L) * (80.0 + 273.15 K) = V2 * (9.0 + 273.15 K)
(1.75×10^3 L) * (353.15 K) = V2 * (282.15 K)
617,852.5 L·K = V2 * 282.15 K

Next, we divide both sides of the equation by 282.15 K to isolate V2:

V2 = (617,852.5 L·K) / (282.15 K)

Performing the division, we get:

V2 ≈ 2,190.58 L

Therefore, the new volume (V2) of the gas is approximately 2,190.58 L.