Question

A student increases the temperature of a 200 cm3 balloon from 33degrees C to 47degrease C. What will the new volume of the ballon be?

Answer 1

The volume is directly related to the temperature of a gas. You can solve this question using the ideal gas equation(specifically Charles' law), but remember to convert the temperature from Celcius to Kelvin.

V1/T1= V2/T2
200 cm3/ (33+273K)= V2/(47+273K)
V2= 200cm3 * 320/306= 209.15cm3

Answer 2

The new volume of balloon when the temperature increases to 47 degree C  is 209$cm^3$

Explanation:

According to temperature-volume law, It states that the volume of a gas is directly proportional to the Kelvin temperature, while pressure and amount are held constant.

The equation is   $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

Where V is volume and T is temperature in Kelvins. We know that  

V1= $200 \mathrm{cm}^{3}$

T1= $33^{\circ} \mathrm{C}$ +273=306 K  

T2= $47^{\circ} \mathrm{C}$+273=320 K  

To find V2, after Rearranging.  

V2=  $\frac{V_{1} T_{2}}{T_{1}}$

V2  =  $\frac{(200 \times 320 \mathrm{K})}{293 \mathrm{K}}$=  209 $cm^3$