Question

combined gas law
a gas is contained in a thick - walled balloon. the system is manipulated,
resulting in the following changes.
the pressure changes from 1.21 atm to 2.52 atm.
the volume changes from 3.75 l to 1.72 l.
the temperature changes from 293 k to an unknown temperature.
what is the new temperature of the system?
280 k
2.57 k
610 k
570 k

Explanation:

Step1: Recall Combined Gas Law

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

Step2: Rearrange for $T_2$

$$T_2 = T_1 \times \frac{P_2V_2}{P_1V_1}$$

Step3: Substitute values

$$T_2 = 293\,\text{K} \times \frac{2.52\,\text{atm} \times 1.72\,\text{L}}{1.21\,\text{atm} \times 3.75\,\text{L}}$$

Step4: Calculate numerator/denominator

Numerator: $2.52 \times 1.72 \approx 4.3344$; Denominator: $1.21 \times 3.75 \approx 4.5375$
Ratio: $\frac{4.3344}{4.5375} \approx 0.955$

Step5: Compute $T_2$

$$T_2 \approx 293 \times 0.955 \approx 280\,\text{K}$$

Answer:

280 K