8g of O2 gas at STP is expanded so that volume is doubled. Thus, work done is
Answers
If 8g of O2 gas at STP is expanded so that volume is doubled then the work done during this process is 135.52 cal.
Explanation:
The mass of oxygen, m = 8 g
The molecular mass of oxygen, M = 32 g/mol
∴ The no. of moles of O₂ = m/M = 8/32 = ¼ moles
We know that at STP,
1 mole of a gas occupies 22.4 litres
∴ ¼ moles of O₂ will occupy = 22.4 × ¼ = 5.6 Litres
It is given that 8 g of O₂ at STP doubles its volume during the expansion, so, we have
V₁ = 5.6 litres and V₂ = 2 × 5.6 = 11.2 litres
Now, we know that
Work done during this process is given by, W = P(V₂ – V₁)
Here, at STP, the standard pressure of the gas, P = 1 atm
Substituting the values in the work done formula, we get
W = 1 x (11.2 – 5.6)
⇒ W = 5.6 L atm
∵ 1 L atm = 24.20 cal
⇒ W = 5.6 * 24.20
⇒ W = 135.52 cal
Thus, the work done is 135.52 cal.
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