Answer: Temperature of one liter of phosphine gas with a mass of 1.725 g and gas pressure of 0.9910 atm is [tex]-36^0C[/tex]
Explanation:
According to the ideal gas equation:
PV=nRT
P = Pressure of the gas = 0.9910 atm
V= Volume of the gas= 1.0 L
T= Temperature of the gas in kelvin = ?
R= Gas constant =[tex]0.0821Latm/Kmol[/tex]
n= moles of gas= [tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{1.725g}{34g/mol}=0.051ol[/tex]
[tex]0.9910atm\times 1.0=0.051\times 0.0821\times T[/tex]
[tex]T=236.7K=(236.7-273)^0C=-36.3^0C[/tex]
Thus the temperature (in °C) of one liter of phosphine gas with a mass of 1.725 g and gas pressure of 0.9910 atm is [tex]-36.3^0C[/tex]
