Question

A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius, what will the volume of the balloon be after he heats it to a temperature of 250 degrees celsius?

Answer 1

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

$PV=nRT$

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 0.4 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

0.4 liter = $0.4\times10^{-3}=4\times10^{-4} m^{3}$

And initial temperature of balloon, $T_{1}$ = 20°C = (273 + 20)K

                                                                          = 293 K

Let the final volume of balloon is $V_{2}$

And a given, final temperature of balloon, $T_{2}$ is 250°C = (273 + 250)K

                                                                                          = 523 K

Now, $V_{1}$ = $KT_{1}$

          $4\times10^{-4}=K\times293\ (equation\ 1 )$

$V_{2}$ = $KT_{2}$

    $=K\times523\ (equation 2)$

Dividing equation 1 and 2,

 $\frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}$

K cancelled by K.

By cross multiplication:

$293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\ = \frac{2092\times10^{-4}}{293} \\ =7.14\times10^{-4}m^{3}$

Now convert it into liter with the help of calculation done above.

$7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter$

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.