Amass of 156.7g of helium gas at an initial temperature of 35.73°c and at an initial absolute pressure of 3.55 atm undergoes an isothermal expansion until its volume increases by a factor of 1.75. (a) what is the final pressure? (b) how much work is done on the gas? (c) how much heat does the gas absorb? (d) what is the change in the total internal energy of the gas? (a) pa answer part (a) (b) joules answer part (b) joules joules answer part (c) answer part (d) (d) submit

Explanation:

Moles of helium ( n ) = 156.7 / 4 = 39.175

Temperature T₁ = 35.73 +273 = 308.73 K

Volume V₁ = V

Pressure P₁ = 3.55 atm

V₂ =1.75 V

a ) For isothermal change

P₁ V₁ = P₂V₂

P₂ = P₁ V₁ / V₂

= 3.55 X V / 1.75 V

= 2.03 atm.

b ) Work done by the gas = nRT ln(V₂/V₁)

= 39.175 X 8.321 X 308.73 X ln 1.75

= 56318.8 J

Work done on the gas = - 56318.8 J

c ) Since there is no change in temperature , internal energy of gas is constant

Q = ΔE + W

ΔE  = 0

Q = W

Work done by gas = heat absorbed

heat absorbed = 56318.8 J

d ) Change in the internal energy of gas is zero because temperature is constant.

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