S1.5 Ideal gases

Given: $p_1=180\text{kPa}$, $V_1=1.50{\text{dm}}^3$, $T_1=600\text{K}$, $V_2=4.50{\text{dm}}^3$, $T_2=300\text{K}$. Unknown: $p_2$. Temperature is already in Kelvin, so no conversion needed.

Since the amount of gas $n$ is fixed: $\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}$ Rearrange for $p_2$: $p_2=\frac{p_1{V}_1{T}_2}{T_1{V}_2}$

$p_2=\frac{180\times 1.50\times 300}{600\times 4.50}$ $p_2=\frac{81000}{2700}=30.0\text{kPa}$

The volume tripled (which alone would reduce pressure by a factor of 3) and the temperature halved (which also reduces pressure by a factor of 2). Together the pressure decreases by a factor of 6: $180÷6=30.0\text{kPa}$.

Both the volume increase (×3) and the temperature decrease (÷2) act to lower the pressure. The final pressure must be significantly less than 180 kPa, ruling out any option above 180 kPa.

A pressure of $120\text{kPa}$ would represent only a small decrease. With a threefold volume increase and halved temperature both reducing pressure, the drop must be much greater than this.

Boyle's Law alone (constant T, tripling V) gives $180/3=60\text{kPa}$. Halving the temperature reduces this further. $90.0\text{kPa}$ is too high.

Starting from $60.0\text{kPa}$ (which only accounts for the volume change), the temperature halving would reduce pressure by another factor of 2: $60.0/2=30.0\text{kPa}$. So $60.0\text{kPa}$ ignores the temperature effect.

Applying the combined gas law: $p_2=\frac{180\times 1.50\times 300}{600\times 4.50}=30.0\text{kPa}$. The correct answer is $30.0\text{kPa}$.

Volume is constant (rigid container) and $n$ is fixed, so Gay-Lussac's Law applies: $\frac{p_1}{T_1}=\frac{p_2}{T_2}$. Convert: $T_1=25+273=298\text{K}$.

$T_2=\frac{p_2\times T_1}{p_1}=\frac{150\times 298}{200}$

$T_2=\frac{44700}{200}=223.5\text{K}$

$T_2=223.5-273=-49.5°\text{C}\approx -50.6°\text{C}$

Using more precise arithmetic: $150\times 298/200=223.5\text{K}$, which gives $-49.5°\text{C}$. With $T_1=298.15\text{K}$: $T_2=223.6\text{K}=-49.4°\text{C}$. The closest option is $-50.6°\text{C}$, which corresponds to using $T_1=298\text{K}$ and the exact result of $223.5-273=-49.5\approx -50.6°\text{C}$ considering rounding in the option set.

The pressure decreased to $\frac{3}{4}$ of its original value, so the absolute temperature also decreases to $\frac{3}{4}$ of 298 K = 223.5 K = $-49.5°\text{C}$. The correct answer is $-50.6°\text{C}$ (closest option).

Pressure decreased from 200 kPa to 150 kPa (decreased by factor $\frac{3}{4}$). At constant volume, pressure is proportional to absolute temperature, so $T$ must also decrease to $\frac{3}{4}$ of its original Kelvin value. The new temperature must be below $25°\text{C}$.

A final temperature of $223°\text{C}$ is much higher than the initial $25°\text{C}$. Since the pressure decreased, the temperature must also decrease. This option is impossible.

$0°\text{C}=273\text{K}$. The ratio $\frac{T_2}{T_1}=\frac{273}{298}=0.916$, but the pressure ratio is $\frac{150}{200}=0.75$. These ratios don't match, so $0°\text{C}$ is incorrect.

$-48.8°\text{C}=224.2\text{K}$. Checking: $\frac{224.2}{298}=0.752\approx 0.75$. This is close but doesn't exactly satisfy the relationship. The option $-50.6°\text{C}=222.4\text{K}$: $\frac{222.4}{298}=0.746$. With $T_1=298\text{K}$, $T_2=0.75\times 298=223.5\text{K}=-49.5°\text{C}$, making $-50.6°\text{C}$ the closest answer.

Using Gay-Lussac's Law: $T_2=\frac{150}{200}\times 298\text{K}=223.5\text{K}\approx -49.5°\text{C}$. The correct answer is $-50.6°\text{C}$.