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What is the temperature of 0.750 mol of a gas stored in a 6,850 mL cylinder at 2.21 atm? Use PV=nRT and R=0.0821 (L•atm)/(mol•K)

A.)2.95 K
B.)5.24 K
C.)138 K
D.)246 K

User Nmock
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2 Answers

3 votes

Answer:

D. 246k

Step-by-step explanation:

Edg

User Nuicca
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Using the formula PV=nRT
we know:
n= 0.750mol
R= 0.0821 (L•atm)/(mol•K)
V= 6850 ml = 6.85 L
P= 2.21 atm
and we need to find T
all you have to do is divide PV by nR and you will find T, like this:

t = (p * v)/(n * r) \\ t = (2.21atm * 6.85l)/(0.750mol * 0.0821 (atm * l)/(mol * k) ) \\ t = 246k
The temperature of the gas is D)246 K
User Gannonbarnett
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