Can you explain the answer of this question below:8 g of O2 gas at STP is expanded so that volume is doubled. Thus, work done isA:22.4 L atmB:11.2L atmC:5.6L atmD:- 5.6L atmThe answer is d.

Question

Answer

Let atomic weight of x = M_x
atomic weight of y = M_y

we know,
mole = weight /atomic weight
a/c to question,

mole of xy₂ = 0.1
so,
0.1 = 10g/( M_x +2M_y)

M_x + 2M_y = 100g -------(1)

for x₃y₂ ; mole of x₃y₂ = 0.05

0.05 = 9/( 3M_x + 2M_y )

3M_x + 2M_y = 9/0.05 = 9 × 20 = 180 g ---(2)
solve eqns (1) and (2)

2M_x = 80
M_x = 40g/mol

and M_y = 30g/mole

Answer

We know that we=-p(v2-v1) as volume is not mentioned,we first find volume 32g of O2=22.4 Lt at STP therefore , 8g=22.4*8/32=5.6lt since the q says,volume is doubled,v=5.6*2=11.2lt p=1 ATM w= -1(11.2-5.6)=-5.6lt atm

Answer

Understanding the Problem
When 8 g of O2 gas is expanded to double its volume at Standard Temperature and Pressure (STP), we need to calculate the work done during this expansion.
Key Concepts
- Work Done in Gas Expansion: The work done (W) on or by the system during an isothermal expansion can be expressed as:
W = -P_ext * ΔV
where P_ext is the external pressure and ΔV is the change in volume.
- STP Conditions: At STP, 1 mole of an ideal gas occupies 22.4 L. The molar mass of O2 is approximately 32 g/mol. Thus, 8 g of O2 corresponds to:
n = 8 g / 32 g/mol = 0.25 moles
- Initial Volume (V1): At STP, the volume of 0.25 moles of O2 is:
V1 = n * 22.4 L = 0.25 * 22.4 L = 5.6 L
- Final Volume (V2): Upon doubling the volume:
V2 = 2 * V1 = 2 * 5.6 L = 11.2 L
Calculating ΔV
- Change in Volume (ΔV):
ΔV = V2 - V1 = 11.2 L - 5.6 L = 5.6 L
Calculating Work Done
- Assuming P_ext at STP: The external pressure (P_ext) at STP is approximately 1 atm.
- Work Done (W):
W = -P_ext * ΔV = -1 atm * 5.6 L = -5.6 L atm
Conclusion
The negative sign indicates that work is done by the system (the gas is doing work on the surroundings). Therefore, the correct answer is D: -5.6 L atm.

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