Question

 In a flexible-walled container, 300 L of gas is prepared at 600 mm Hg and 200 oC. The gas is placed in a tank under high pressure. When the tank cools to 20 oC, the pressure of the gas is 30 atm. What is the new volume of the gas?

Answer 1

Initial Pressure $P_1$ = 600mm Hg = 600/760 atm = 0.7894 atm

Initial volume $V_1$ = 300L

Initial temperature $T_{1}$ = 473K

Final temperature $T_2$ = 293K

Final Pressure $P_2$ = 30 atm

Since, moles of gas remains constant in both the cases

So, $\dfrac{P_1V_1}{RT_1}= \dfrac{P_2V_2}{RT_2}$

So, final volume of the gas will be $V_2=\dfrac{P_1V_1T_2}{T_1P_2}$

Substituting the values $V_2= \dfrac{0.7894\times300\times293}{473\times30}=4.89 L$

Final Volume of the gas is 4.89L