Answer :
Answer:
Part A: [tex]\mathbf{Q =94 \ J}[/tex] to two significant figures
Part B: [tex]\mathbf{Q =160 \ J}[/tex] to two significant figures
Part C: [tex]\mathbf{Q =220 \ J}[/tex] to two significant figures
Explanation:
Given that :
mass of the hydrogen = 0.30 g
the molar mass of hydrogen gas molecule = 2 g/mol
we all know that:
number of moles = mass/molar mass
number of moles = 0.30 g /2 g/mol
number of moles = 0.15 mol
For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = [tex]C_v=\dfrac{3}{2}R[/tex]
For Part A:
[tex]Q = mC_v\Delta T[/tex]
[tex]Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K[/tex]
[tex]Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)[/tex]
[tex]Q=93.5325 \ J[/tex]
[tex]\mathbf{Q =94 \ J}[/tex] to two significant figures
Part B. For hot temperature, [tex]C_v=\dfrac{5}{2}R[/tex]
[tex]Q = mC_v\Delta T[/tex]
[tex]Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K[/tex]
[tex]Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)[/tex]
[tex]Q=155.8875 \ J[/tex]
[tex]\mathbf{Q =160 \ J}[/tex] to two significant figures
Part C. For an extremely hot temperature, [tex]C_v=\dfrac{7}{2}R[/tex]
[tex]Q = mC_v\Delta T[/tex]
[tex]Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K[/tex]
[tex]Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)[/tex]
[tex]Q=218.2425 \ J[/tex]
[tex]\mathbf{Q =220 \ J}[/tex] to two significant figures
Part A) Heat needed to change the temperature of the gas from 50 K to 100 K is 94J.
Part B) Heat needed to change the temperature of the gas from 250 K to 300 K is 160J.
Part C) Heat needed to change the temperature of the gas from 2250 K to 2300 K is 220J.
What is heat?
The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.
Given is the mass of the hydrogen = 0.30 g and the molar mass of hydrogen gas molecule = 2 g/mol
Then, number of moles = mass/molar mass
number of moles = 0.30 g /2 g/mol = 0.15 mol
Part A) For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule Cv =3/2R
Q =m Cv ΔT
Q = 0.15 mol x 3/2 x 8.314 x (100 - 50)
Q = 93.5325 Joules.
Q = 94 J to two significant figures.
Part B) For hot temperature, Cv = 5/2 R
Heat needed, Q =m Cv ΔT
Q = 0.15 mol x 5/2 x 8.314 x (300 - 250)
Q = 155.8875Joules.
Q = 160 J to two significant figures.
Part C) In case of extremely hot temperature, Cv =7/2R
Heat needed to change the temperature from 2250 K to 2300 K
Q = 0.15 mol x 7/2 x 8.314 x (2300 - 2250)
Q = 218.2425 J
Q = 220J
Learn more about heat.
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