Question

A 12.0 L sample of gas is at STP. What would be its new volume if its pressure was decreased to 575 mmHg and its temperature was doubled?

Answer 1

`V_2=31.7L`

Explanation:

Let us first list the given data:

Position 1:
STP means that `P=1atm=760mmHg` and `T=273K`.
The volume occupied by the gas is `V=12.0L`.

Position 2:
The new conditions are:
`P=575mmHg` and `T=2xx273=546K`.
`V=?`

To find the volume of the gas at position 2, we can rearrange the ideal gas law: `PV=nRT` to be:

`(PV)/T=k` where `k=nR`, which is constant.

Therefore, `(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

`=>V_2=(P_1V_1)/(T_1)xx(T_2)/(P_2)=(760cancel(mmHg)xx12.0L)/(273cancel(K))xx(546cancel(K))/(575cancel(mmHg))=31.7L`