Question

g If 6.35 moles of a monatomic ideal gas at a temperature of 320 K are expanded isothermally from a volume of 1.45 L to a volume of 3.95 L, calculate: a) the change in the internal energy of the gas. b) the work done by the gas. c) the heat flow into or out of the gas.

Answer 1

Answer:

(a)  change in the internal energy of the gas is zero

(b) the work done by the gas is 16.93 kJ

(c) the heat flow is 16.93 kJ, which is into the gas

Explanation:

Given;

number of moles of gas, n = 6.35 moles

temperature of the gas, T = 320 K

initial volume of the gas, V₁ = 1.45 L

final volume of the gas, V₂ = 3.95 L

Part (a)

For isothermal expansion, temperature is constant and internal energy will also be constant.

Therefore, change in the internal energy of the gas is zero since the gas expanded isothermally (constant temperature).

ΔU = Q - W

where;

ΔU is change in internal energy

Q is heat transferred to the system

W is the work done by the system

Thus, Q = W

ΔU = 0

Part (b)

the work done by the gas

$W = nRTln{[\frac{V_2}{V_1}]$

where;

R is gas constant = 8.314 J/mol.K

$W = (6.35)(8.314)(320)ln{[\frac{3.95}{1.45}]}\\\\W =16930.4\ J\\\\W = 16.93\ kJ$

Part (c)

the heat flow into or out of the gas

Q = ΔU + W

Q = 0 + 16.93 kJ

Q = 16.93 kJ

Since the heat flow is positive, then it is heat flow into the gas.