Question #131621

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?

Expert's answer

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, P1V1=P2V2P_1V_1=P_2V_2

And V2=P1V1P2V_2=\dfrac{P_1V_1}{P_2}

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

Thus, P1=715  mmHg×1  atm760  mmHg=0.941  atmP_1=715\;\rm mmHg\times \dfrac{1\;\rm atm}{760\;\rm mmHg}=0.941\;\rm atm

P2=640  mmHg×1  atm760  mmHg=0.842  atmP_2=640\;\rm mmHg\times \dfrac{1\;\rm atm}{760\;\rm mmHg}=0.842\;\rm atm

And the initial volume in L is V1=600  mL×1  L1000  mL=0.600  LV_1=600\;\rm mL\times \dfrac{1\;L}{1000\;mL}=0.600\;L

V2=0.941  atm×0.600  L0.842  atm=0.671  LV_2=\dfrac{0.941\;\rm atm\times 0.600\;L}{0.842\;\rm atm}=0.671\;\rm L

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