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A toy balloon is initially at a temperature of T = 16o C. Its initial volume is 0.0042 m3 (It is a 10 cm radius sphere). Assume that the pressure in the balloon always equals atmospheric pressure, 101.3 kPa.

1)

The balloon is lying in the sun, which causes the volume to expand by 9%. What is the new temperature, T?

T =

o C

2)

By how many Joules did the internal energy of the gas in the balloon increase? Air is almost entirely diatomic gas.

ΔU =J

3)

How much work was done by the gas during the expansion? (It's a positive number.)

Wby =

J

4)

How much heat was added to the gas by the Sun?

Q =

Respuesta :

Answer:

Given that

T₁=16⁰C

V₁=0.0042 m³

P=101.3 KPa

1)

Assuming that gas is ideal gas.

At constant pressure

[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]

V₂=1.09 V₁

T₁=16⁰C=289 K

By putting the values

[tex]\dfrac{1.09V_1}{1V_1}=\dfrac{T_2}{289}[/tex]

T₂=315.01 K

T₂=42.01⁰C

2)

The internal energy for air

ΔU= m Cv ΔT

P V = m RT ( ideal gas equation)

[tex]m=\dfrac{101.3\times 0.0042}{0.287\times 289}\ kg[/tex]

m=0.0051 kg

Cv for air = 0.71 KJ/kg.k

By putting the values

ΔU= m Cv ΔT

[tex]\Delta U=0.0051\times 0.71\times (42.01-16)\ KJ[/tex]

ΔU=0.094 KJ

3)

V₁=0.0042 m³

V₂=1.09 V₁

V₂=0.0045 m³

Work W

W= P.ΔV

W= 101.3 x (0.0045-0.0042)

W=0.038 KJ

4)

We know that

From first law of thermodynamics

Q= ΔU+ W

By putting the value

Q= 0.038+0.094

Q=0.132 KJ

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