Question

Two litre of $N_2$ at $0^{o}C$ and 5 atm are expanded isothermally against a constant external pressure of 1 atm until the pressure of gas reaches 1 atm. Assuming the gas to be ideal calculate work of expansion?

• A. $-504.2joule$
• B. $-405.2joule$
• C. $+810.4joule$
• D. $-810.4joule$

Answer 1

The correct option is A $-504.2joule$

$\text{Given:-}{P}_1=5atm$

$P_2=1atm$

$V_2=2litres$

$\text{Solution:-}$

Initially the gas is at 5 atm pressure and it's volume is 2 liter. The final pressure is going to be decreased to 1 atm. Process is isothermal means the temperature is constant. Using Boyle's law we calculate the final volume of the gas:

$P_1{V}_1=P_2{V}_2$

Let's plug in the values in the equation to calculate final volume:

$5atm\times \left(2L\right)=1atm.\left(V_2\right)$

$V_2=10liters$

External pressure is also given as 1 atm. Pressure-volume work is calculated using the formula:

$w=-P_{ext}\Delta V$

$w=-1atm(10L-5L)$

$w=-5atm\cdot L$

since, $1atm\cdot L=101.325J$

So, $-5atm\cdot L\left(\frac{101.325J}{1atm\cdot L}\right)$

$=-504.2J$

$\text{Hence the correct option is A}$