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0.0625 mol of an ideal gas is contained within a 1.0000 Lvolume container. Its pressure is 142,868 Pa and temperature is X K . What is X?

(HINT: |X| is near an order of magnitude of 102 K ).

1 Answer

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Answer: The temperature of the ideal gas is
2.75* 10^2K

Step-by-step explanation:

To calculate the temperature, we use the equation given by ideal gas equation:


PV=nRT

where,

P = Pressure of the gas = 142,868 Pa = 142.868 kPa (Conversion factor: 1 kPa = 1000 Pa)

V = Volume of gas = 1.0000 L

n = number of moles of ideal gas = 0.0625 moles

R = Gas constant =
8.31\text{L kPa }mol^(-1)K^(-1)

T = temperature of the gas = ?

Putting values in above equation, we get:


142.868kPa* 1.0000=0.0625mol* 8.31\text{L kPa }mol^(-1)K^(-1)* T\\\\T=275K=2.75* 10^2K

Hence, the temperature of the ideal gas is
2.75* 10^2K

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