At what temperature (in C) will a sample of gas occupy 91.3 L if it occupies 45.0 L at 70.0°C? Assume constant pressure.)
Answers
Solution is here,
for initial case,
temperature(T1)=70°C=70+ 273=343K
vloume( V1) =45 L
for final case,
temperature( T2)=?
volume(V2)= 91.3 L
at constant pressure,
V1/V2 = T1/T2
or, 45/91.3 = 343/ T2
or, T2= (343×91.3)/45
or, T2=695.9 K = (695.9-273)°C=422.9°C
Final answer:
The temperature needed for a sample of gas to occupy 91.3 L, when originally it occupied 45.0 L at 70.0°C at constant pressure, is about 423.71°C.
Explanation:
This question pertains to the ideal gas law equation (P1V1/T1 = P2V2/T2, where P is pressure, V is volume, and T is temperature), we must first convert our temperatures to Kelvin (K) since the ideal gas law uses absolute temperature. To convert from degrees Celsius to K, add 273.15, thus 70.0°C becomes 343.15K. As the problem states pressure is constant, we can ignore that component of the equation and it simplifies to V1/T1 = V2/T2.
By rearranging the equation to solve for V2, we get V2 = V1 (T2/T1). Given the volume of the gas at the initial temperature (V1 = 45.0 L), and temperature T2 with unknown value, we rewrite the equation as T2 = (V2 * T1) / V1. Substituting the given values into the equation gives T2 = (91.3 L * 343.15K) / 45.0 L, which calculates to approximately 696.86K, converting back to Celsius gives us around 423.71°C.
So, your sample of gas would need to be heated to approximately 423.71°C in order for it to occupy a volume of 91.3 L at constant pressure.
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your answer is b.
hope this helps!!!