Consider five moles of helium (a monatomic gas) under the ambient conditions (P

Question

Consider five moles of helium (a monatomic gas) under the ambient conditions (P = 1.013 × 105 Pa and T = 293.15 K). The gas constant is R = 8.3145 J/(molK).

(12) What is the volume of the gas?
(A) 0.120 m3; (B) 0.130 m3; (C) 0.140 m3; (D) 0.150 m3;
(E) 0.160 m3.

(13) What is the internal energy of the gas?
(A) 14.3 kJ; (B) 15.3 kJ; (C) 16.3 kJ; (D) 17.3 kJ; (E) 18.3 kJ.

(14) What is the work done by the gas if it expands to twice of its
volume adiabatically?
(A) 3.76 kJ; (B) 4.76 kJ; (C) 5.76 kJ; (D) 6.76 kJ; (E) 7.76 kJ.

Explanation / Answer

here,

number of moles , n = 5

P = 1.013 * 10^5 Pa

T = 293.15 K

R = 8.314

1)

let the volume of the gas be V

for an ideal gas

P*V = n * R * T

1.013 * 10^5 * V = 5 * 8.314 * 293.15

V= 0.12 m^3

the volume of the gas is (A) 0.12 m^3

2)

the internal energy of the gas , U = 1.5 *n * R * T

U = 1.5 * 5 * 8.314 * 293.15

U = 18.3 KJ

the internal energy of the gas is (E) 18.3 KJ

3)

W = For adiabatic process;

(T'/T) = (V/V')^(k-1)

k = 1.4 for ideal gas.

T'/T = (1/2)^0.4 = 0.757

T' = 0.757* 293.15 = 221.91 K

work: W = (k/1-k) R(T' - T)

W = (1.4/(-0.4)) (8.314)(221.91 - 293.15)

W = 2114 J

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