Question

A 0.133 mol sample of gas in a 525 ml container has a pressure of 312 torr. the temperature of the gas is ________ °
c.

Answer 1

Assuming the gas is an ideal gas, we use the equation PV = nRT where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the universal gas constant.

PV = nRT
312/760 (.525) = 0.133(0.08205)T
T = 20.21 K

Answer 2

Answer: The temperature of the gas is -253.25°C

Explanation:

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

$PV=nRT$

where,

P = pressure of the gas = 312 torr

V = Volume of the gas = 525 mL = 0.525 L  (Conversion factor: 1 L = 1000 mL)

T = Temperature of the gas

R = Gas constant = $62.364\text{ L.Torr }mol^{-1}K^{-1}$

n = number of moles of gas = 0.133 mole

Putting values in above equation, we get:

$312Torr\times 0.525L=0.133mol\times 62.364\text{ L.Torr }mol^{-1}K^{-1}\times T\\\\T=\frac{312\times 0.525}{62.364\times 0.133}=19.75K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$19.75K=T(^oC)+273\\T(^oC)=(19.75-273)=-253.25^oC$

Hence, the temperature of the gas is -253.25°C