The surface tension of the soap water solution is (1/10 π) N / m. The free energy of the surface layer of a soap bubble of diameter 5 mm will be

Question

The surface tension of the soap water solution is (1/10 π) N / m. The free energy of the surface layer of a soap bubble of diameter 5 mm will be (A) 2.5 × 10-6 J (B) 1 × 10-7 J (C) 8 × 106 J (D) 5 × 10-6 J

Tardigrade Admin · Accepted Answer

As, we know that the surface energy of a soap bubble,

W = A T

So,

W^{'} = 2 A^{'} T [A^{'} = 2 A]

Since, a soap bubble has double surface area.
Given, surface tension of soap water solution,

T = \frac{1}{10 π} N / m

and diameter of soap bubble

= 5 mm

Now,

W^{'} = 2 (π D^{2}) T = 2 \times π \times 25 \times 1 0^{- 6} \times \frac{1}{10 π}

\Rightarrow W^{'} = 5 \times 1 0^{- 6} J

Q. The surface tension of the soap water solution is $\frac{1}{10 π} N / m$ . The free energy of the surface layer of a soap bubble of diameter $5 mm$ will be

$2.5 \times 1 0^{- 6} J$

$1 \times 1 0^{- 7} J$

$8 \times 1 0^{6} J$

$5 \times 1 0^{- 6} J$

Q. The surface tension of the soap water solution is 10π1​N/m. The free energy of the surface layer of a soap bubble of diameter 5mm will be

2.5×10−6J

1×10−7J

8×106J

5×10−6J

As, we know that the surface energy of a soap bubble, W=AT So, W′=2A′T[A′=2A] Since, a soap bubble has double surface area. Given, surface tension of soap water solution, T=10π1​N/m and diameter of soap bubble =5mm Now, W′=2(πD2)T=2×π×25×10−6×10π1​ ⇒W′=5×10−6J

Q. The surface tension of the soap water solution is $\frac{1}{10 π} N / m$ . The free energy of the surface layer of a soap bubble of diameter $5 mm$ will be

$2.5 \times 1 0^{- 6} J$

$1 \times 1 0^{- 7} J$

$8 \times 1 0^{6} J$

$5 \times 1 0^{- 6} J$