Answer:
V₂ = 1070 mL or 1.07 L
Solution:
Data Given;
P₁ = 1170 mmHg
V₁ = 915 mL
T₁ = 24 °C + 273 K = 297 K
P₂ = 842 mmHg
V₂ = ?
T₂ = - 23 °C + 273 K = 250 K
According to Ideal gas equation,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = P₁ V₁ T₂ / P₂ T₁
Putting Values,
V₂ = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)
V₂ = 1070 mL or 1.07 L
<span>The ideal gas law.
PV=nRT
pressure x volume = moles x Faraday's constant x Temp Kelvin (C+273)
Original data
Pressure 1 atmosphere
Volume 1 liter
Temp 25C = 298K
New data
Volume 0.5 liter
pressure X
Temp 260C = 533K
P1v1T1 = P2v2T2
plug and chug.
(1)(1)(293) = (x)(0.5)(533)
Solve for X, which is the new pressure. </span>
H, my day has been alright, how’s yours?
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit.
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust. Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress.
I hope this helped! These are COMPLEX questions though! =D
