An ideal gas expands in volume from $1 \times 10^{- 3}$ 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

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What is the amount of work done when two moles of ideal gas is compressed from a volume of 1m3 to 10dm3 at 300 K against a pressure of 100 kPa?

The correct option is A 99 kJV1=1m3,Pressure =100 kPa =100000 Pa V2=10dm3=(1/100)m3. So, W=PΔV=100000×(V2−V1) =105×(1/100−1) =99000J=99 kJ.

What is the amount of work done when two moles of ideal gas is compressed from a volume of $1 m^{3}$ to $10 d m^{3}$ at 300 K against a pressure of 100 kPa?

The correct option is A $99$ kJ
$V_{1} = 1 m^{3}$ ,
Pressure $= 100$ kPa $= 100000$ Pa
$V_{2} = 10 d m^{3} = (1 / 100) m^{3}$ .
So, $W = P Δ V = 100000 \times (V_{2} - V_{1})$
$= 10^{5} \times (1 / 100 - 1)$
$= 99000 J = 99$ kJ.