Respuesta :
Answer:
Explanation:
Atmospheric pressure P = 10⁵ N / m ²
change in volume Δ V = 16 - 12 = 4 m³
work done by gas against constant pressure
W = P Δ V
Putting the values above
work done = 10⁵ x 4
W = 4 x 10⁵ J .
B ) Heat added Q = 254 x 10³ cals
= 254 x 4.2 x 10³ J
= 10.66 x 10⁵ J
According to first law of thermodynamics
Q = Δ E + W
Q is heat energy added , Δ E is increase in internal energy and W is work done by gas.
Putting the values in the equation above
10.66 x 10⁵ = Δ E +4 x 10⁵
Δ E = 6.66 x 10⁵ J .
A gas is enclosed in a cylinder maintained at atmospheric pressure absorbs 254 kcal of heat and its volume increases from 12.0 m³ to 16.2 m³. The work done by the gas is -1.02 × 10⁵ kcal and the change in the internal energy is -1.01 × 10⁵ kcal.
A gas enclosed in a cylinder maintained at atmospheric pressure (P = 101325 Pa) has a volume increase from 12.0 m³ to 16.2 m³.
We can calculate the work done by the gas (w) using the following expression.
[tex]w = -P \times \Delta V = -101,325Pa \times (16.2m^{3}-12.0m^{3} )= -4.26 \times 10^{5} J \times \frac{1kcal}{4184J} = -1.02 \times 10^{5} kcal[/tex]
254 kcal of heat is added to the gas, so q = 254 kcal. We can calculate the change in the internal energy (ΔE) using the following expression.
[tex]\Delta E = q + w = 254 kcal + (-1.02 \times 10^{5} kcal) = -1.01 \times 10^{5} kcal[/tex]
A gas is enclosed in a cylinder maintained at atmospheric pressure absorbs 254 kcal of heat and its volume increases from 12.0 m³ to 16.2 m³. The work done by the gas is -1.02 × 10⁵ kcal and the change in the internal energy is -1.01 × 10⁵ kcal.
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