Answer:
A) T1 = 566 k = 293°C
B) T2 = 1132 k = 859°C
Explanation:
A)
The average kinetic energy of the molecules of an ideal gas is givwn by the formula:
K.E = (3/2)KT
where,
K.E = Average Kinetic Energy
K = Boltzman Constant
T = Absolute Temperature
At 10°C:
K.E = K10
T = 10°C + 273 = 283 K
Therefore,
K10 = (3/2)(K)(283)
FOR TWICE VALUE OF K10:
T = T1
Therefore,
2 K10 = (3/2)(K)(T1)
using the value of K10:
2(3/2)(K)(283) = (3/2)(K)(T1)
<u>T1 = 566 k = 293°C</u>
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B)
The average kinetic energy of the molecules of an ideal gas is given by the formula:
K.E = (3/2)KT
but K.E is also given by:
K.E = (1/2)(m)(vrms)²
Therefore,
(3/2)KT = (1/2)(m)(vrms)²
vrms = √(3KT/m)
where,
vrms = Root Mean Square Velocity of Molecule
K = Boltzman Constant
T = Absolute Temperature
m = mass
At
T = 10°C + 273 = 283 K
vrms = √[3K(283)/m]
FOR TWICE VALUE OF vrms:
T = T2
Therefore,
2 vrms = √(3KT2/m)
using the value of vrms:
2√[3K(283)/m] = √(3KT2/m)
2√283 = √T2
Squaring on both sides:
(4)(283) = T2
<u>T2 = 1132 k = 859°C</u>
Answer:
heat required in pan B is more than pan A
Explanation:
Heat required to raise the temperature of the substance is given by the formula

now we know that both pan contains same volume of water while the mass of pan is different
So here heat required to raise the temperature of water in Pan A is given as


Now similarly for other pan we have


So here by comparing the two equations we can say that heat required in pan B is more than pan A
Answer:
d ≈ 7,6 g/cm³
Explanation:
d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³
V = l³ = (1.74cm)³ ≈ 5,27 cm³
Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that

T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C