Answer:
Approximately [tex]2.2\; \rm cm^{3}[/tex] (assuming that this gas is ideal, and that the initial volume of this gas is [tex]4.0\; \rm cm^{3}[/tex].)
Explanation:
In this question, both the pressure on the gas and the temperature of the gas have changed. However, the ideal gas laws (Boyle's Law and Charles' Law) requires that only one of the two quantity change at a time. Therefore, consider this change in two steps:
- Change pressure from [tex]P_0 = 2.0 \times 10^{5}\; \rm Pa[/tex] to [tex]P_1= 1.6 \times 10^{5}\; \rm Pa[/tex] while keeping temperature constant at [tex]288\; \rm K[/tex]. Find the new volume [tex]V_1[/tex] using Boyle's Law.
- After that, keep the pressure the same (at [tex]1.6 \times 10^{5}\; \rm Pa[/tex]) and change the temperature from [tex]T_1 = 288\; \rm K[/tex] to [tex]T_2 = 125\; \rm K[/tex]. Find the final volume of the gas [tex]V_2[/tex] using Charles' Law.
Boyle's Law states that for a fixed quantity of some ideal gas, if temperature is held constant, the volume of the gas will be inversely proportional to the pressure on the gas.
Let [tex]V_0[/tex] denote the initial volume of this gas. The question states that at [tex]P_0 = 2.0\times 10^{5}\; \rm Pa[/tex] and [tex]T_0 = 288\; \rm K[/tex], the volume of the gas is [tex]V_0 = 4.0\; \rm cm^{3}[/tex].
By Boyle's Law, if temperature is held constant ([tex]T_1 = T_0 = 288\; \rm K[/tex],) then at [tex]P_1 = 1.6 \times 10^{5}\; \rm Pa[/tex]:
[tex]\begin{aligned} V_1 &= V_0 \cdot \frac{P_0}{P_1} \\ &= 4.0\; \rm cm^3 \times \frac{2.0 \times 10^{5}\; \rm Pa}{1.6 \times 10^{5}\; \rm Pa} \approx 5.0\; \rm cm^{3} \end{aligned}[/tex].
On the other hand, Charles' Law suggests that for a fixed quantity of some ideal gas, if the pressure of the gas is held constant, the volume of the gas will be proportional to the temperature (in degree Kelvins) of the gas.
Let [tex]V_1[/tex] denote the volume of this gas before the temperature change. At [tex]P_0 = 2.0\times 10^{5}\; \rm Pa[/tex] and [tex]T_0 = 288\; \rm K[/tex], previous calculations show that [tex]V_0 = 5.0\; \rm cm^{3}[/tex].
By Charles' Law, if the pressure of this gas is held constant ([tex]P_2 = P_1 =1.6 \times 10^{5}\; \rm Pa[/tex],) then at the new temperature [tex]T_2 = 125\; \rm K[/tex]:
[tex]\begin{aligned} V_2 &= V_1 \cdot \frac{T_1}{T_0} \\ &= 5.0\; \rm cm^3 \times \frac{125\; \rm K}{288\; \rm K} \approx 2.2\; \rm cm^{3} \end{aligned}[/tex].
Therefore, at [tex]T_2 = 125\; \rm K[/tex] and [tex]P = 1.6 \times 10^{5}\; \rm Pa[/tex], the volume of this gas would be approximately [tex]2.2\; \rm cm^{3}[/tex].
