Question

One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and volume of 25.0 L. Determine:

a) the initial volume ?
b) the temperature of the gas?
(Note: 1 atm = 1.01 x 105Pa, universal gas constant R = 8.31 J/mol K, 1 L = 10-3m3)

Answer 1

Answer:

(a) Initial volume will be 7.62 L

(b) Final temperature will be 303.85 K

Explanation:

We have given one mole of ideal gas done 3000 J

So work done W = 3000 J

Let initial volume is $V_1$ and initial pressure $P_1=1atm$ ( As pressure is constant )

Final volume $V_2=25L$ = 0.025 $m^3$

Number of moles n = 1

(B) From ideal gas of equation we know that $PV=nRT$

So $1.01\times 10^5\times0.025=1\times 8.31\times T$

T = 303.85 Kelvin

(B) For isothermal process work done is equal to

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

$3000=1\times 8.314\times 303.85\times ln\frac{0.025}{V_1}$

$ln\frac{0.025}{V_1}=1.1881$

$\frac{0.025}{V_1}=3.2808$

$V_2=0.00846m^3=7.62L$

So initial volume will be 7.62 L

Answer 2

(a) the initial volume of the gas is 8.46 L

(b) the temperature of the gas is 303.85 K

Isothermal expansion:

Given that the gas expands isothermally.

The number of moles of the gas is n = 1

The final pressure of the gas P' = 1 atm

The final volume of the gas is V' = 25L

The ideal gas equation is given by:

PV = nRT

Let the final temperature be T', so:

P'V' = nRT'

1.01 × 10⁵ Pa × 25×10⁻³m³ = 1 × 8.31 J/mol.K × T'

T' = 303.85K

Since the expansion is isothermal, the temperature of the gas does not change, therefore the initial temperature is also T'.

Initial temperature T = 303.85K

The work done by the ideal gas is given by:

dW = PdV

$dW=\frac{nRT}{V}dV\\\\dW=nRT\ln[V]^{V'}_{V}\\\\W=nRT\ln\frac{V'}{V}\\\\ln\frac{V'}{V}=\frac{W}{nRT}\\\\\ln\frac{0.025}{V}=\frac{3000}{1\times8.31times303.85}$

V = 0.00846m³

V = 8.46L is the initial volume.

Learn more about isothermal expansion:

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