## College Physics (4th Edition)

The new pressure of the gas is $1.62~atm$
We can find an expression for $v_0$: $\overline{KE} = \frac{3}{2}~k~T$ $\frac{1}{2}m~v_0^2 = \frac{3}{2}~k~T$ $v_0 = \sqrt{\frac{3~k~T}{m}}$ $v_0 = \sqrt{\frac{3~P~V}{N~m}}$ We can find the new pressure when the rms speed is $0.90~v_0$: $0.90~v_0 = 0.90~\sqrt{\frac{3~P~V}{N~m}}$ $0.90~v_0 = \sqrt{\frac{3~(0.90)^2~P~V}{N~m}}$ $0.90~v_0 = \sqrt{\frac{3~(0.81~P)~V}{N~m}}$ We can find the new pressure of the gas: $0.81~P = (0.81)(2.0~atm) = 1.62~atm$ The new pressure of the gas is $1.62~atm$.