Question

sample a
mass of each particle: 17 u
average particle speed: 2,100 m/s
which sample has the higher temperature?
sample a
sample b
neither; the samples have the same temperature
which sample has the higher gas pressure?
sample a
sample b
neither; the samples have the same gas pressure

Explanation:

Step1: Recall kinetic - theory formula

The average kinetic energy of a gas particle is $\overline{K}=\frac{3}{2}kT$, and also $\overline{K}=\frac{1}{2}m\overline{v^{2}}$. For temperature comparison, we equate $\frac{3}{2}kT=\frac{1}{2}m\overline{v^{2}}$, so $T = \frac{m\overline{v^{2}}}{3k}$.
For sample A, $m_A = 17u$, $\overline{v_A}=2100m/s$, then $T_A=\frac{17\times(2100)^{2}}{3k}$.
For sample B, $m_B = 2u$, $\overline{v_B}=2000m/s$, then $T_B=\frac{2\times(2000)^{2}}{3k}$.
Calculate $T_A=\frac{17\times4410000}{3k}=\frac{74970000}{3k}$ and $T_B=\frac{2\times4000000}{3k}=\frac{8000000}{3k}$. Since $T_A>T_B$, sample A has a higher temperature.

Step2: Recall pressure - kinetic theory relation

The pressure of a gas is given by $P=\frac{2}{3}n\frac{1}{2}m\overline{v^{2}}$, where $n$ is the number density of particles. Assuming the number density $n$ is the same for both samples (not given otherwise).
$P_A=\frac{2}{3}n\times\frac{1}{2}m_A\overline{v_A^{2}}=\frac{1}{3}n\times17\times(2100)^{2}$ and $P_B=\frac{1}{3}n\times2\times(2000)^{2}$.
$P_A=\frac{1}{3}n\times17\times4410000=\frac{74970000n}{3}$ and $P_B=\frac{1}{3}n\times2\times4000000=\frac{8000000n}{3}$. Since $P_A > P_B$, sample A has a higher pressure.

Answer:

Sample A has the higher temperature.
Sample A has the higher gas pressure.