At a constant pressure p, the plot of volume (V) as a function of temperature (T) for 2 moles of an ideal gas gives a straight line with a slope 0.328 LK-14. The value of p (in atm) is closest to [Gas constant, R = 0.0821 L atm mol-1 K-1]

Question

At a constant pressure p, the plot of volume (V) as a function of temperature (T) for 2 moles of an ideal gas gives a straight line with a slope 0.328 LK-14. The value of p (in atm) is closest to [Gas constant, R = 0.0821 L atm mol-1 K-1] (A) 0.25 (B) 0.5 (C) 1.0 (D) 2.0

Tardigrade Admin · Accepted Answer

According to ideal gas equation pV=nRT ⇒ (V/T)=(nR/p)=slope Given, slope = 0.328, n = 2 ∴ p=(nR/slope)=(2×0.0821/0.328)=0.500 atm

Q. At a constant pressure $p$ , the plot of volume $(V)$ as a function of temperature $(T)$ for $2$ moles of an ideal gas gives a straight line with a slope $0.328 L K^{- 1} 4$ . The value of p (in atm) is closest to [Gas constant, $R = 0.0821 L$ atm $m o l^{- 1} K^{- 1}$ ]

0.25

0.5

1.0

2.0

According to ideal gas equation
$p V = n RT \Rightarrow \frac{V}{T} = \frac{n R}{p} =$ slope
Given, slope $= 0.328, n = 2$
$∴ p = \frac{n R}{s l o p e} = \frac{2 \times 0.0821}{0.328} = 0.500$ atm

Q. At a constant pressure p, the plot of volume (V) as a function of temperature (T) for 2 moles of an ideal gas gives a straight line with a slope 0.328LK−14. The value of p (in atm) is closest to [Gas constant, R=0.0821L atm mol−1K−1]

0.25

0.5

1.0

2.0

According to ideal gas equation pV=nRT⇒TV​=pnR​=slope Given, slope =0.328,n=2 ∴p=slopenR​=0.3282×0.0821​=0.500 atm

Q. At a constant pressure $p$ , the plot of volume $(V)$ as a function of temperature $(T)$ for $2$ moles of an ideal gas gives a straight line with a slope $0.328 L K^{- 1} 4$ . The value of p (in atm) is closest to [Gas constant, $R = 0.0821 L$ atm $m o l^{- 1} K^{- 1}$ ]

According to ideal gas equation
$p V = n RT \Rightarrow \frac{V}{T} = \frac{n R}{p} =$ slope
Given, slope $= 0.328, n = 2$
$∴ p = \frac{n R}{s l o p e} = \frac{2 \times 0.0821}{0.328} = 0.500$ atm