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Q.

When 1 gm of water at 0°C and $1 \times 10^{5} N / m^{2}$ pressure is converted into ice of volume 1.091 cm², the external work done will be:

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a

0.0091 joule

b

0.0182 joule

c

– 0.0091 joule

d

– 0.0182 joule

answer is A.

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Detailed Solution

It is an isothermal process. Hence work done

$= P (V_{2} - V_{1}) = 1 \times 10^{5} \times (1.091 - 1) \times 10^{- 6} = 0.0091 J$

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