A weather balloon filled with He gas has a volume of \(2.00 \times 10^{3} \mathrm{m}^{3}\) at ground level, where the atmospheric pressure is 1.000 atm and the temperature \(27^{\circ}\) C. After the balloon rises high above Earth to a point where the atmospheric pressure is 0.340 atm, its volume increases to \(5.00 \times 10^{3} \mathrm{m}^{3} .\) What is the temperature of the atmosphere at this altitude?

Short Answer

Expert verified
The temperature of the atmosphere at the altitude where the atmospheric pressure is 0.34 atm is approximately \( -18.04^{\circ}C\)

Step by step solution

01

Initial Temperature Conversion

Convert the initial temperature from Celsius to Kelvin by using the conversion equation \(K = C + 273.15\). Thus, the initial temperature in Kelvin is \(27^{\circ}C + 273.15K = 300.15K\).
02

Apply the Ideal Gas Law

Since the number of moles and the gas constant don't change, and we can equate the initial and final states of the balloon using the equation \(P_1V_1/T_1 = P_2V_2/T_2\), where 1 refers to the initial state and 2 refers to the final.
03

Solve for Final Temperature

We can rearrange the equation from step 2 to solve for \(T_2\) giving us the equation \(T_2 = P_2V_2T_1/(P_1V_1)\). Substituting the given values into this equation gives us the final temperature in Kelvin as \(T_2 = (0.340\, atm \times 5000\,m^3 \times 300.15\,K) / (1.000\, atm \times 2000\,m^3) = 255.1125\,K\).
04

Final Temperature Conversion

Convert the final temperature from Kelvin back to Celsius using the conversion equation \(C = K - 273.15\), giving a final temperature of \(255.1125\,K - 273.15 = -18.04^{\circ}C.\)

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