Answer to Question #131621 in Molecular Physics | Thermodynamics for Angel

Question

A gas occupies a volume of 600 ml at 715 mmHg. If the pressure is decreased to 640 mmHg at constant temperature, what is the resulting volume of the gas?

Answer 1

According to the ideal gas law PV=nRT, where P is the pressure in atm, V is the volume in L, n is the amount of moles, R is the gas constant, and T is the temperature in K

The right side of the equation is constant, since the amount of gas doesn't change and change in pressure occurs at constant temperature

Thus, $P_1V_1=P_2V_2$

And $V_2=\dfrac{P_1V_1}{P_2}$

To convert mmHg to atm, divide the value in mmHg to 760

Thus, $P_1=715\;\rm mmHg\times \dfrac{1\;\rm atm}{760\;\rm mmHg}=0.941\;\rm atm$

$P_2=640\;\rm mmHg\times \dfrac{1\;\rm atm}{760\;\rm mmHg}=0.842\;\rm atm$

And the initial volume in L is $V_1=600\;\rm mL\times \dfrac{1\;L}{1000\;mL}=0.600\;L$

$V_2=\dfrac{0.941\;\rm atm\times 0.600\;L}{0.842\;\rm atm}=0.671\;\rm L$