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