Question

The pressure of a gas remains constant while the temperature, volume, and internal energy of the gas increase by 57.7 Co, 1.24 x 10⁻³ m³, and 843 J, respectively. The mass of the gas is 21.6 g, and its specific heat capacity (at constant pressure) is 1080 J/(kg Co). Determine the pressure.

Answer 1

Answer:

P = 405.67 kPa

Explanation:

given,

change in temperature = 57.7 °C

volume of the gas = 1.24 x 10⁻³ m³

Internal energy of the = 843 J

Cp = 1080 J/Kg.C°

mass = 21.6 g

heat capacity

Q = m Cp ΔT

Q = 0.0216 x  1080 x 57.7

Q = 1346.025 J

according to first law of thermodynamics

U = Q + W

W = Q - ΔU

PΔV = Q - ΔU

PΔV = 1346.025 - 843

P x  1.24 x 10⁻³  = 503.0256

P = 405.67 x 10³ Pa

P = 405.67 kPa

The pressure of the gas is equal to P = 405.67 kPa

Answer 2

The pressure when the  mass of the gas is 21.6 g, and its specific heat capacity (at constant pressure) is 1080 J/(kg Co) should be  405.67 kPa.

Calculation of the pressure:

Since

change in temperature = 57.7 °C

volume of the gas = 1.24 x 10⁻³ m³

Internal energy of the gas = 843 J

Cp = 1080 J/Kg.C°

mass = 21.6 g

So, we know that

Q = m Cp ΔT

Q = 0.0216 x  1080 x 57.7

Q = 1346.025 J

Now we know that

U = Q + W

So

W = Q - ΔU

So,

PΔV = Q - ΔU

PΔV = 1346.025 - 843

P x  1.24 x 10⁻³  = 503.0256

P = 405.67 x 10³ Pa

P = 405.67 kPa

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