The sample has a new pressure of 274kPa. If at 105 kPa and 275K, a 220 mL sample of helium gas is contained in a cylinder with a moving piston. The sample is pushed till it has a 95.0 mL volume and 310K .
The macroscopic characteristics of ideal gases are related by the ideal gas law (PV = nRT). A gas is considered to be perfect if its particles (a) do not interact with one another and (b) occupy no space (have no volume). Where P= pressure V= volume and T = temperature.
From ideal gas equation
P₁V₁/T₁ =P₂V₂/T₂
105×220÷275 = P₂ ×95÷310
P₂= (105×220×310)÷(275×95)
P2= 7161000/26125
P2 = 274.105 kPa
Hence, the new pressure of helium gas is 274kPa
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The answer is a because a fill in the blank is always just one word
Explanation:
Both cohesion and molecular interchange contribute to liquid viscosity. The impact of increasing the temperature of a liquid is to reduce the cohesive forces while simultaneously increasing the rate of molecular interchange. The former effect causes a decrease in the shear stress while the latter causes it to increase.
temperature?
The viscosity of liquids decreases rapidly with an increase in temperature, and the viscosity of gases increases with an increase in temperature. Thus, upon heating, liquids flow more easily, whereas gases flow more sluggishly.
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Explanation:
The lack of a control group
The easiest way is to use the Law of Gay-Lussac. This law states that there is a direct relation between the temperature in Kelvin of a gas and the pressure.
Then, namig p the pressure and T the temperature in Kelvin and using subscripts for every state:
p/T is constant ==> p_1 / T_1 = p_2/T_2
From which you obtain:
p_2 = [p_1 / T_1] * T_2
T_1 = 33.0 + 273.15 = 306.15 K
T _2 = 21.4 + 273.15 = 294.55 K
p_1 = 1014 kPa
p_2 = 1014 kPa * 294.55 K / 306.15 K = 975.6 kPa
