At text27 texto textC one mole of an ideal gas exerted a pressure of 0.821 atmosphere. What is its volume in liters ( textR = 0 text.0821 textL atm textmo textl-1K-1 ?

Question

At text27 texto textC one mole of an ideal gas exerted a pressure of 0.821 atmosphere. What is its volume in liters ( textR = 0 text.0821 textL atm textmo textl-1K-1 ? (A) 300 (B) 30 (C) 0.3 (D) 3

Tardigrade Admin · Accepted Answer

T=27+273=300K P=0.821 textatm V=? R=0.0821 textLatm textmo textl-1 textK-1 n=1 mol PV=nRT V=(nRT/P) =(1× 0.0821× 300/0.821)=30 L

Q. At $27^{o} C$ one mole of an ideal gas exerted a pressure of 0.821 atmosphere. What is its volume in liters ( $R = 0 .0821 L$ atm $mo l^{- 1} K^{- 1}$ ?

300

30

0.3

3

$T = 27 + 273 = 300 K$ $P = 0.821 atm$ $V = ?$ $R = 0.0821 Latm mo l^{- 1} K^{- 1}$ $n = 1 m o l$ $P V = n RT$ $V = \frac{n RT}{P}$ $= \frac{1 \times 0.0821 \times 300}{0.821} = 30 L$

Q. At 27oC one mole of an ideal gas exerted a pressure of 0.821 atmosphere. What is its volume in liters ( R = 0.0821L atm mol−1K−1 ?

300

30

0.3

3

T=27+273=300K P=0.821atm V=? R=0.0821Latmmol−1K−1 n=1mol PV=nRT V=PnRT​ =0.8211×0.0821×300​=30L

Q. At $27^{o} C$ one mole of an ideal gas exerted a pressure of 0.821 atmosphere. What is its volume in liters ( $R = 0 .0821 L$ atm $mo l^{- 1} K^{- 1}$ ?

$T = 27 + 273 = 300 K$ $P = 0.821 atm$ $V = ?$ $R = 0.0821 Latm mo l^{- 1} K^{- 1}$ $n = 1 m o l$ $P V = n RT$ $V = \frac{n RT}{P}$ $= \frac{1 \times 0.0821 \times 300}{0.821} = 30 L$