$2 k g$ of water is converted into steam by boiling at atmospheric pressure. The volume changes from $2 \times 10^{- 3} m^{3}$ to $3.34 m^{3}$ . The work done by the system is about

- 340 k J

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- 170 k J

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170 k J

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340 k J

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Solution

The correct option is D $340 k J$
$Given: P_{a} = 1.013 \times 10^{5} p a, V_{1} = 2 \times 10^{- 3} m^{3}, V_{2} = 3.34 m^{3}$

and workdone is given by :

$W = P Δ V$

Putting values:

$W = 1.013 \times 10^{5} \times (3.34 - 2 \times 10^{- 3}) = 340 K J$

Hence option D is correct.

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Similar questions

Two kg of water is converted into steam by boiling at atmospheric pressure. The volume changes from $2 \times 10^{- 3} m^{3}$ to $3.34 m^{3}$ The work done by the system is about

Let 1.00 kg of liquid water at $100^{\circ} C$ be converted to steam at $100^{\circ} C$ by boiling at standard atmospheric pressure (which is 1.00 atm or 1.01 $\times$ 105 Pa) in the arrangement as shown. The volume of that water changes from an initial value of $1.00 \times 10^{- 3}$ $m^{3}$ as a liquid to 1.671 $m^{3}$ as steam.

1. How much work is done by the system during this process?

What is the change in the system's internal energy during the process?

2. If the latent heat of vaporization is 2256 kJ/kg, how much energy is transferred as heat during the process?

Q. What is the free energy change

(△ G)

when 1.0 mole of water at

100^{o} C

and 1 atmospheric pressure is converted into steam at

100^{o} C

and 1 atmospheric pressure?

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2 kg of water is converted into steam by boiling at atmospheric pressure. The volume changes from 2×10−3 m3 to 3.34 m3. The work done by the system is about

The correct option is D 340 kJ Given: Pa=1.013×105pa,V1=2×10−3 m3, V2=3.34 m3and workdone is given by :W=PΔVPutting values:W=1.013×105×(3.34−2×10−3)=340KJHence option D is correct.

$2 k g$ of water is converted into steam by boiling at atmospheric pressure. The volume changes from $2 \times 10^{- 3} m^{3}$ to $3.34 m^{3}$ . The work done by the system is about

The correct option is D $340 k J$
$Given: P_{a} = 1.013 \times 10^{5} p a, V_{1} = 2 \times 10^{- 3} m^{3}, V_{2} = 3.34 m^{3}$

and workdone is given by :

$W = P Δ V$

Putting values:

$W = 1.013 \times 10^{5} \times (3.34 - 2 \times 10^{- 3}) = 340 K J$

Hence option D is correct.