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If 3.67 mol of an ideal gas has a pressure of 2.70 atm and a volume of 74.51 l, what is the temperature of the sample in degrees celsius

a) 321.4 °C
b) 268.7 °C
c) 305.2 °C
d) 292.3 °C

User Zsuzsa
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1 Answer

7 votes

Final answer:

Using the ideal gas law, the temperature of the ideal gas sample was calculated to be 392.45 °C. However, this temperature does not match any of the given options, indicating a possible error in the provided choices.

The correct answer is none of above.

Step-by-step explanation:

To calculate the temperature of the sample in degrees Celsius, we will use the ideal gas law which is PV = nRT, where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin. First, we solve for T by rearranging the equation: T = PV / (nR). Here we know that P = 2.70 atm, V = 74.51 L, and n = 3.67 mol. The value of R to use is 0.0821 L·atm/(mol·K) when pressure is in atm and volume is in liters.

T = (2.70 atm * 74.51 L) / (3.67 mol * 0.0821 L·atm/(mol·K))

T = 200.577 atm·L / (0.301387 mol·K)

T = 665.6 K

To convert Kelvin to Celsius, we subtract 273.15 from the Kelvin temperature.

T in Celsius = 665.6 K - 273.15 = 392.45 °C

However, it appears there might be an error with the given options, as none of them match the calculated temperature. The correct temperature of the sample is 392.45 °C.

User Adamax
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