The volume versus temperature graph of 1 mol | vedclass

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(A) According to the ideal gas equation,$pV = nRT$. For $1 \ mole$ of gas,$n = 1$,so $pV = RT$ or $p = \frac{RT}{V}$. Taking $R = 0.0821 \ L \ atm \ K

Vedclass · Accepted Answer

$0.328, 0.820, 0.820$

Vedclass · Answer

(A) According to the ideal gas equation,$pV = nRT$. For $1 \ mole$ of gas,$n = 1$,so $pV = RT$ or $p = \frac{RT}{V}$. Taking $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$. For point $X$: $V = 50 \ L, T = 200 \ K$. $p_X = \frac{0.0821 	imes 200}{50} = 0.3284 \ atm \approx 0.328 \ atm$. For point $Y$: $V = 50 \ L, T = 500 \ K$. $p_Y = \frac{0.0821 	imes 500}{50} = 0.821 \ atm \approx 0.820 \ atm$. For point $Z$: $V = 20 \ L, T = 200 \ K$. $p_Z = \frac{0.0821 	imes 200}{20} = 0.821 \ atm \approx 0.820 \ atm$. Thus,the pressures at $X, Y, Z$ are $0.328 \ atm, 0.820 \ atm, 0.820 \ atm$ respectively.

The volume versus temperature graph of $1 \ mole$ of an ideal gas is given below. The pressure of the gas (in $atm$) at $X, Y$ and $Z$,respectively,are

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