An ideal gas expands in volume from 1- 10-3m3 to 1- 10-2 m3 at 300 K against a constant pressure of 1- 105 Nm-2- The work done is-

The correct option is A 900 J Given, V_1=1× 10^ 3m^3 V_2=1× 10^ 2m^3 P=1× 10^5Nm^ 2 Work done, W= P(V_2 V_1) = 1× 10^5(1× 10^ 2 1× ...

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Standard XII

Chemistry

Mole Concept

An ideal gas ...

Question

An ideal gas expands in volume from $1 \times 10^{- 3} m^{3}$ to $1 \times 10^{- 2} m^{3}$ at $300 K$ against a constant pressure of $1 \times 10^{5} N m^{- 2}$ .

The work done is:

- 900 J

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- 900 k J

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270 k J

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900 k J

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Solution

The correct option is A $- 900 J$
Given,
$V_{1} = 1 \times 10^{- 3} m^{3}$
$V_{2} = 1 \times 10^{- 2} m^{3}$
$P = 1 \times 10^{5} N m^{- 2}$
Work done, $W = - P (V_{2} - V_{1}) = - 1 \times 10^{5} (1 \times 10^{- 2} - 1 \times 10^{- 3})$
$⟹ W = - 900 J$
Hence option A is the right answer.

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An ideal gas expands in volume from 1×10−3m3 to 1×10−2 m3 at 300 K against a constant pressure of 1×105 Nm−2. The work done is:

The correct option is A −900 JGiven,V1=1×10−3m3V2=1×10−2m3P=1×105Nm−2Work done, W=−P(V2−V1)=−1×105(1×10−2−1×10−3)⟹W=−900JHence option A is the right answer.

An ideal gas expands in volume from $1 \times 10^{- 3} m^{3}$ to $1 \times 10^{- 2} m^{3}$ at $300 K$ against a constant pressure of $1 \times 10^{5} N m^{- 2}$ .

The work done is:

The correct option is A $- 900 J$
Given,
$V_{1} = 1 \times 10^{- 3} m^{3}$
$V_{2} = 1 \times 10^{- 2} m^{3}$
$P = 1 \times 10^{5} N m^{- 2}$
Work done, $W = - P (V_{2} - V_{1}) = - 1 \times 10^{5} (1 \times 10^{- 2} - 1 \times 10^{- 3})$
$⟹ W = - 900 J$
Hence option A is the right answer.